Toán 10 $\color{Blue}{\fbox{Topic Hot}\bigstar\text{Thảo Luận Về Bất Đẳng Thức}\bigstar}$

[viethoang1999]

72) Cho $a;b;c>0$ thỏa $a+b+c=3$. Cmr: $\sum \sqrt{1+a^2+2bc}\le 6$
73) Cho $a;b;c\ge 0$ thỏa $a+b+c=3$.Cmr: $\sum \sqrt{a}\ge \sum ab$
74) Cho $a;b;c>0$ thỏa: $ab^2+bc^2+ca^2=3$. Cmr: $\sum \sqrt[3]{a+7}\le 2\sum a^4$
75) Cho $x;y;z>0$ thỏa: $\sum \dfrac{1}{1+x}\ge 2$.
Tìm Max $B=xyz$
Tổng quát: Cho $a_1;a_2;...;a_n>0$ thỏa: $\dfrac{1}{1+a_1}+\dfrac{1}{1+a_2}+...+\dfrac{1}{1+a_n}\ge n-1$.
Tìm Max $A=a_1.a_2...a_n$

76) Tìm Max $A=\dfrac{x^4+x+1+32\sqrt[4]{x^3-4x^2+7x-12}}{x^4+x^2+16x-11}$

77) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=1$. Cmr: $\sum \dfrac{a^2}{1+b-a}\ge 1$

78) Cho $x;y;z>0$ thỏa $xy^2z^2+x^2z+y=3z^2$. Tìm Max $Q=\dfrac{z^4}{1+z^4(x^4+y^4)}$

79) Cho $a;b;c>0$. Cmr: $\sum \dfrac{bc}{a^2+2bc}\le 1$

80) Cho $a;b;c;d>0$. Cmr: $\sum \dfrac{a^3}{a^2+b^2}\ge \dfrac{a+b+c+d}{2}$

 

[vipboycodon]

Dạo này pic vắng quá nhỉ.
75) Ta có:
$\dfrac{1}{x+1} \ge 1-\dfrac{1}{y+1}+1-\dfrac{1}{z+1} = \dfrac{y}{y+1}+\dfrac{z}{z+1} \ge 2\sqrt{\dfrac{yz}{(y+1)(z+1)}}$ (cauchy)
Tương tự ta có: $\dfrac{1}{y+1} \ge 2\sqrt{\dfrac{xz}{(x+1)(z+1)}}$
$\dfrac{1}{z+1} \ge 2\sqrt{\dfrac{xy}{(x+1)(y+1)}}$
Nhân vế với vế đc:
$\dfrac{1}{(x+1)(y+1)(z+1)} \ge \dfrac{8xyz}{(x+1)(y+1)(z+1)}$
<=> $xyz \le \dfrac{1}{8}$

 

[vipboycodon]

73) $\sqrt{a}+\sqrt{b}+\sqrt{c} \ge ab+bc+ac = \dfrac{(a+b+c)^2-a^2-b^2-c^2}{2} = \dfrac{9-(a^2+b^2+c^2)}{2}$
<=> $a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 9$
Theo bdt cauchy ta có:
$a^2+\sqrt{a}+\sqrt{a} \ge 3a$
$b^2+\sqrt{b}+\sqrt{b} \ge 3b$
$c^2+\sqrt{c}+\sqrt{c} \ge 3c$
=> $a^2+b^2+c^2+2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \ge 3(a+b+c) = 9$
Vậy BĐT ban đầu đúng.
Dấu "=" xảy ra khi $a = b = c = 1$

 

[forum_]

72/

$\sum_{cyc} \sqrt{1+a^2+2bc}$ \leq $\sqrt{3\sum_{cyc}(1+a^2+2bc)}=\sqrt{3(3+9)}=6$

76/

Đạo hàm, tìm đc Max = 1 khi x=4

78/

Đặt $x=a$, $y=b$ and $\dfrac{1}{z}=c$.

\Rightarrow $a^2c+b^2a+c^2b=3$.

Do đó: $\left(\dfrac{a^4+b^4+c^4}{3}\right)^3$ \geq $\left(\dfrac{a^3+b^3+c^3}{3}\right)^4$ \geq $\left(\dfrac{a^2c+b^2a+c^2b}{3}\right)^4=1$.

\Leftrightarrow $a^4+b^4+c^4$ \geq $3$.

Dấu "=" khi $a=b=c=1$ or $x=y=z=1$.

Do đó $Q$ \leq $\dfrac{1}{3}$.

79/

$\sum_{cyc}\dfrac{bc}{a^2+2bc}$ \leq $1$ \Leftrightarrow $\sum_{cyc}\left(\dfrac{bc}{a^2+2bc}-\dfrac{1}{2}\right)$ \leq $1-\dfrac{3}{2}$ \Leftrightarrow $\sum_{cyc}\dfrac{a^2}{a^2+2bc}$ \geq $1$

80/

Vì $\dfrac{a^3}{a^2+b^2}$ \geq $a-\dfrac{b}{2}$

\Leftrightarrow..... \Leftrightarrow.....

77/

$\sum \dfrac{a^2}{1+b-a}$ \geq 1$

\Leftrightarrow $\sum \dfrac{a^4}{a^2+a^2b-a^3}$ \geq 1

C-S ......

$a^3+b^3+c^3$ \geq $a^2b+b^2c+c^2a$ (đúng)

 

[viethoang1999]

76)
Cách 2:
$x^3-4x^2+7x-12=(x-3)(x^2-x+4)$
ĐKXĐ: $\left\{\begin{matrix}x\ge 3 & & \\ x^4+x^2+16x-11\ne 0 & & \end{matrix}\right.$
Có: $32\sqrt[4]{x^3-4x^2+7x-12}=4\sqrt[4]{16.16.16.(x-3)(x^2-x+4)}\le 16+16+16x-48+x^2-x+4=x^2+15x-12$
\Rightarrow $x^4+x+1+32\sqrt[4]{x^3-4x^2+7x-12}\le x^4+x^2+16x-11$
\Rightarrow $A\le 1$

$"="$ \Leftrightarrow $x=4$
78)
Tắt, làm lại:
Đặt $\dfrac{1}{z}=t$

$Q=\dfrac{1}{x^4+y^4+t^4}$
Từ $GT$\Rightarrow $xy^2+yt^2+tx^2=3$
$9=(xy^2+yt^2+tx^2)^2\le (x^2+y^2+t^2)(x^4+y^4+t^4)\le \sqrt{3(x^4+y^4+t^4)}(x^4+y^4+t^4)$
Vậy $x^4+y^4+t^4\le 3$
\Rightarrow $Q\le \dfrac{1}{3}$
$"="$ \Leftrightarrow $a=b=c=\dfrac{1}{3}$

79)
Cách 1: Làm như bài của forum_, bước cuối đúng theo BCS dạng cộng mẫu (Cauchy Schwarz dạng Engel)
Cách 2: $\dfrac{a^2+2bc}{bc}=\dfrac{a^2}{bc}+2\ge \dfrac{2a^2}{b^2+c^2}+2=\dfrac{2(a^2+b^2+c^2)}{b^2+c^2}$
$\dfrac{a^2+2bc}{bc}=\dfrac{a^2}{bc}+2\ge \dfrac{2a^2}{b^2+c^2}+2=\dfrac{2(a^2+b^2+c^2)}{b^2+c^2}$ \Rightarrow $\dfrac{bc}{a^2+2bc}\le \dfrac{b^2+c^2}{2\sum a^2}$
\Leftrightarrow $\sum \dfrac{bc}{a^2+2bc}\le 1$
$"="$ \Leftrightarrow $a=b=c>0$

80)
Sử dụng Cauchy ngược dấu ta được:
$\sum \dfrac{a^3}{a^2+b^2}=\sum \dfrac{a(a^2+b^2)-ab^2}{a^2+b^2}=\sum a-\sum \dfrac{ab^2}{a^2+b^2}\ge \sum a-\sum \dfrac{ab^2}{2ab}=\sum a-\sum \dfrac{b}{2}=\dfrac{a+b+c+d}{2}$
$"="$ \Leftrightarrow $a=b=c=d>0$

 

[viethoang1999]

77)
Cách 2:
Từ $GT$ \Rightarrow $a;b;c\in (0;1)$
\Rightarrow $\left\{\begin{matrix}1+b-a>0 & & \\ 1+c-b>0 & & \\ 1+a-c>0 \end{matrix}\right.$
Có: $1\ge 1-(a-b)^2=(1-a+b)(1+a-b)$
\Rightarrow $\dfrac{1}{1+b-a}\ge 1+a-b$ \Leftrightarrow $\dfrac{a^2}{1+b-a}\ge a^2+a^2(a-b)$
\Leftrightarrow $\sum \dfrac{a^2}{1+b-a}\ge \sum a^2+\sum a^2(a-b)=1+\sum a^2(a-b)$
Vậy ta cần CM: $\sum a^2(a-b)\ge 0$ \Leftrightarrow $\sum a^3-\sum a^2b\ge 0$
Theo AM-GM thì $a^3+a^3+b^3\ge 3a^2b$, tương tự cộng lại ta có đpcm.
$"="$ \Leftrightarrow $a=b=c=\dfrac{1}{\sqrt{3}}$

Cách 3:

$\sum \left [\dfrac{a^2}{1+b-a}+a^2(1+b-a) \right ]\ge 2\sum a^2$
\Rightarrow $\sum \dfrac{a^2}{1+b-c}\ge 2\sum a^2-\sum a^2(1+b-a)=\sum a^2-\sum a^2b+\sum a^3\ge \sum a^2=1$
(Do $\sum a^3\ge \sum a^2b$ đã chứng minh ở cách 2)

 

[viethoang1999]

74) Cho $a;b;c>0$ thỏa: $ab^2+bc^2+ca^2=3$. Cmr: $\sum \sqrt[3]{a+7}\le 2\sum a^4$
75) Cho $x;y;z>0$ thỏa: $\sum \dfrac{1}{1+x}\ge 2$.
Tìm Max $B=xyz$
Tổng quát: Cho $a_1;a_2;...;a_n>0$ thỏa: $\dfrac{1}{1+a_1}+\dfrac{1}{1+a_2}+...+\dfrac{1}{1+a_n}\ge n-1$.
Tìm Max $A=a_1.a_2...a_n$

74)
Áp dụng AM-GM có:
$a^{4}+a^{4}+c^{4}+1\ge 4\sqrt[4]{a^{8}c^{4}}=4a^{2}c$
Tương tự cộng lại:
$3(\sum a^{4}+1)\ge 4(\sum a^{2}c)$
\Rightarrow $\sum a^{4}\ge 3$ $(1)$
Áp dụng AM-GM có:
$\sum \sqrt[3]{a+7}\le \dfrac{1}{12}\sum (a+23)=\dfrac{1}{48}\sum (4a+92)\le \dfrac{1}{48}\sum (a^{4}+95)= \dfrac{1}{48}(\sum a^{4}+285)$ $(2)$
Từ $(1);(2)$ suy ra
$\sum \sqrt[3]{a+7}\le \dfrac{1}{48}(96\sum a^{4})=2\sum a^{4}$

75)
Tổng quát: Làm tương tự...
\Rightarrow $A\le 2^n$
Dấu $"="$ xảy ra khi: $a_1=a_2=...=a_n=\dfrac{1}{2}$

 

[viethoang1999]

Một số bài tập về kĩ thuật Cauchy ngược dấu.

81) Cho $a;b;c;d>0$ thỏa $a+b+c+d=4$. Cmr: $\sum \dfrac{a}{1+b^2}\ge 2$

82) Cho $a;b;c;d>0$ thỏa $a+b+c+d=4$. Cmr: $\sum \dfrac{a}{1+b^2c}\ge 2$

83) Cho $a;b;c>0$ thỏa $ab+bc+ca=3$. Cmr: $\sum \dfrac{a^2}{a+2b^2}\ge 1$

84) Cho $a;b;c>0$ thỏa $a+b+c=3$. Cmr: $\sum \dfrac{a+1}{b^2+1}\ge 3$

85) Cho $a;b;c>0$ thỏa $a+b+c=3$. Cmr: $\sum \dfrac{a+b}{1+a}\ge \sum ab$

 

[hien_vuthithanh]

81/

$\dfrac{a}{1+b^2}$=$a-\dfrac{ab^2}{1+b^2}$\geq $a-\dfrac{ab^2}{2b}$=$a-\dfrac{ab}{2}$
tt \Rightarrow $ \sum \dfrac{a}{1+b^2}$ \geq$ \sum a-\dfrac{\sum ab}{2}$=4-$ \dfrac{\sum ab}{2}$
có ab+bc+cd+da=(a+c).(b+d)\leq $(\dfrac{a+b+c+d}{2})^2$=4
\Rightarrow $ \sum \dfrac{a}{1+b^2}$ \geq 4-$\dfrac{4}{2}$=2
\Rightarrow dpcm

 

[hien_vuthithanh]

82/

$\sum \dfrac{a}{1+b^2c}$=$\sum \dfrac{a^2}{a+ab^2c}$\geq $\dfrac{(\sum a)^2}{\sum a+\sum ab^2c}$=$\dfrac{16}{4+\sum ab^2c}$ @};-
ta có$ab^2c+bc^2d+cd^2a+da^2b$
=(ab+cd)(bc+ad)\leq $(\dfrac{ab+bc+cd+da}{2})^2$=$(\dfrac{(a+b)(c+d)}{2})^2$\leq $(\dfrac{\dfrac{a+b+c+d}{2}}{2})$=4
\Rightarrow $\sum ab^2c $\leq 4
thay vào @};- \Rightarrow $\sum \dfrac{a}{1+b^2c}$ \geq 2

 

[hien_vuthithanh]

84/

$\sum \dfrac{a+1}{b^2+1}$=$\sum\dfrac{a}{b^2+1}$+$\sum \dfrac{1}{b^2+1}$=$\sum(a-\dfrac{ab^2}{b^2+1})$+$\sum(1-\dfrac{b^2}{b^2+1})$
\geq $\sum(a-\dfrac{ab^2}{2b})$+$\sum(1-\dfrac{b^2}{2b})$=$\sum(a-\dfrac{ab}{2})$+$\sum(1-\dfrac{b}{2})$=$\sum a-\sum \dfrac{ab}{2}+3-\sum \dfrac{b}{2}$\geq$\dfrac{9}{2}$-$\dfrac{\dfrac{1}{3}.(\sum a)}{2}$=3

 

[hien_vuthithanh]

85/

$\sum \dfrac{a+b}{1+a}$=$\sum(1-\dfrac{a^2-b}{1+a})$=$\sum a$-$\sum\dfrac{a^2}{1+a}$+$\sum \dfrac{b}{1+a}$ =3-$\sum\dfrac{a^2}{1+a}$+$\sum \dfrac{b}{1+a}$
ta có +/ $\sum \dfrac{a^2}{1+a}$=$\sum a $-$\sum \dfrac{a}{1+a}$ \geq 3-$\dfrac{1}{4}$.$\sum(a+1)$=3-$\dfrac{1}{4}$.6=$\dfrac{3}{2}$ @};-
+/ 3$\sum ab$\leq $(\sum a)^2$=9 \Rightarrow $\sum ab$ \leq 3
+/ $\sum \dfrac{b}{1+a}$=$\sum \dfrac{b^2}{b+ab}$\geq $ \dfrac{(\sum b)^2}{\sum ab+3}$=$\dfrac{9}{\sum ab+3} $\geq $\dfrac{9}{3+3} $=$\dfrac{3}{2} $@};-@};-
Từ @};- và @};-@};- \Rightarrow $\sum \dfrac{a+b}{1+a}$ \geq 3\geq$\sum ab$

 

[viethoang1999]

Một số bài tập về kĩ thuật Cauchy ngược dấu.

82) Cho $a;b;c;d>0$ thỏa $a+b+c+d=4$. Cmr: $\sum \dfrac{a}{1+b^2c}\ge 2$

Đang ở phần bài tập Cauchy ngược dấu, bạn lại dùng Schwarz à!

82)
Cách 2:
$\sum \dfrac{a}{1+b^2c}=\sum \left ( a-\dfrac{ab^2c}{1+b^2c} \right ) \ge \sum \left ( a-\dfrac{ab^2c}{2b\sqrt{c}} \right ) = \sum \left ( a-\dfrac{ab\sqrt{c}}{2} \right ) \ge \sum \left ( a-\dfrac{b\sqrt{a^2c}}{2} \right ) \ge \sum \left ( a-\dfrac{b(a+ac)}{4} \right ) = a+b+c+d-\dfrac{1}{4}(ab+bc+cd+da+abc+bcd+cda+dab)$
Có:
$\bullet $ $ab+bc+cd+da=(a+c)(b+d)\le \dfrac{(a+b+c+d)^2}{4}=4$
$\bullet $ $abc+bcd+cda+dab=ab(c+d)+cd(a+b)\le \dfrac{1}{4}(a+b)^2(c+d)+\dfrac{1}{4}(c+d)^2(a+b)=\dfrac{1}{4}(a+b)(c+d)(a+b+c+d)=(a+b)(c+d)\le \dfrac{(a+b+c+d)^2}{4}=4$
Vậy $\sum \dfrac{a}{1+b^2c}\ge 4-\dfrac{1}{4}(4+4)=2$
$"="$ \Leftrightarrow $a=b=c=d=1$

 

[viethoang1999]

Một số bài tập về kĩ thuật Cauchy ngược dấu.


85) Cho $a;b;c>0$ thỏa $a+b+c=3$. Cmr: $\sum \dfrac{a+b}{1+a}\ge \sum ab$

85)
Cách 2:
$\sum \dfrac{a+b}{1+a}=\sum \left [ a+b-\dfrac{a(a+b)}{1+a} \right ] =2\sum a-\sum \dfrac{a(a+b)}{1+a}$
Có: $(1+a)^2\ge 4a$\Rightarrow $\dfrac{1+a}{4}\ge \dfrac{a}{1+a}$
\Rightarrow $\dfrac{a(a+b)}{1+a}\le \dfrac{(1+a)(1+b)}{4}=\dfrac{\sum a^2+ab+a+b}{4}$\Rightarrow $\sum \dfrac{a(a+b)}{1+a}\le \dfrac{1}{4}\left ( \sum a^2+\sum ab+2\sum a \right ) = \dfrac{1}{4} \left ( (\sum a)^2-\sum ab+6 \right ) =\dfrac{1}{4}\left ( 15-\sum ab \right )$
\Rightarrow $\sum \dfrac{a+b}{1+a}\ge 6-\dfrac{15}{4}+\dfrac{1}{4}\sum ab=\dfrac{9}{4}+\dfrac{1}{4}\sum ab$
Mà $9=(a+b+c)^2\ge 3(ab+bc+ca)$\Rightarrow $ab+bc+ca\le 3$
\Rightarrow $\dfrac{3}{4}(ab+bc+ca)\le \dfrac{9}{4}$
Vậy ta có đpcm.
Cách 3:
$\dfrac{a+b}{1+a}+\dfrac{(a+b)(1+a)}{4}\ge a+b$
\Rightarrow $\sum \dfrac{a+b}{1+a}+\dfrac{\sum a^2+\sum ab+2\sum a}{4}\ge 2\sum a$\Leftrightarrow $\sum \dfrac{a+b}{1+a}+\dfrac{(\sum a)^2+2\sum a}{4}\ge 2\sum a+\dfrac{\sum ab}{4}$
\Leftrightarrow $\sum \dfrac{a+b}{1+a}+\dfrac{15}{4}\ge 6+\dfrac{\sum ab}{4}$
\Leftrightarrow $\sum \dfrac{a+b}{1+a}\ge \dfrac{\sum ab}{4}+\dfrac{9}{4}$

Mà $9=(a+b+c)^2\ge 3(ab+bc+ca)$\Rightarrow $ab+bc+ca\le 3$
\Rightarrow $\dfrac{3}{4}(ab+bc+ca)\le \dfrac{9}{4}$
Vậy ta có đpcm.

 

[viethoang1999]

Một số bài tập về kĩ thuật Cauchy ngược dấu.

83) Cho $a;b;c>0$ thỏa $ab+bc+ca=3$. Cmr: $\sum \dfrac{a^2}{a+2b^2}\ge 1$

83)
Có: $2ab+1=ab+ab+1\ge 3\sqrt[3]{a^2b^2}$ (Cauchy)
\Rightarrow $\dfrac{2}{3}\sqrt[3]{a^2b^2}\le \dfrac{2}{9}(2ab+1)$

Áp dụng AM-GM có:
$\sum \dfrac{a^{2}}{a+2b^{2}}= \sum \left ( a-\dfrac{2ab^{2}}{a+2b^{2}} \right ) \ge \sum \left ( a-\dfrac{2ab^{2}}{3\sqrt[3]{ab^{4}}} \right ) = \sum \left ( a-\dfrac{2\sqrt[3]{a^{2}b^{2}}}{3} \right ) \ge \sum \left ( a-\dfrac{2}{9}(2ab+1) \right ) = a+b+c-\dfrac{4(ab+bc+ac)}{9}-\dfrac{2}{3}$
Ta có
$(a+b+c)^{2}\ge 3(ab+bc+ac)= 9$\Rightarrow $a+b+c\ge 3$
Vậy ta có đpcm.

 

[viethoang1999]

86) Cho $\left\{\begin{matrix}a_1;a_2;...;a_n>0 & & \\ a_1+a_2+...+a_n=1 & & \end{matrix}\right.$. Cmr: $\left ( \dfrac{1}{a_1}-1 \right ) \left ( \dfrac{1}{a_2}-1 \right )...\left ( \dfrac{1}{a_n}-1 \right ) \ge (n-1)^n$

87) Cho $x+y+z=0$. Tìm Min $P=\sum \sqrt{3+4^x}$

88) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=3$. Tìm Min $A=\sum \dfrac{a^3}{\sqrt{b^2+3}}$

89) Cho $a;b;c>0$ thỏa: $a+b+c=\dfrac{3}{4}$. Tìm Min $P=\sum \dfrac{1}{\sqrt[3]{a+3b}}$

90) Cho $x;y;z>0$ thỏa: $xyz=1$. Tìm Min $A=\sum \dfrac{x^2(y+z)}{y\sqrt{y}+2z\sqrt{z}}$

 

[hien_vuthithanh]

88/

A=$\sum \dfrac{a^3}{\sqrt{b^2+3}}$=$\sum \dfrac{a^4}{\sqrt{a^2b^2+3a^2}}$\geq $ \dfrac{(\sum a^2)^2}{\sum (\sqrt{a^2b^2+3a^2)}}$=$ \dfrac{9}{\sum (\sqrt{a^2b^2+3a^2)}}$ &gt;-
AD BCS $(\sqrt{a^2b^2+3a^2}+\sqrt{b^2c^2+3b^2}+\sqrt{c^2c^2+3c^2})^2$\leq 3.$(\sum(a^2b^2)+3\sum a^2)$=$3\sum(a^2b^2)+27$ @};-
ta có $a^2b^2+b^2c^2$=$(b^2+1)(a^2+c^2)-(a^2+c^2)$\leq $(\dfrac{\sum a^2+1}{2})^2$-$(a^2+c^2)$=4-$(a^2+c^2)$
tt \Rightarrow $2\sum (a^2b^2)$\leq 12-2$\sum a^2$=12-6=6 \Rightarrow$3 \sum (a^2b^2)$\leq 9 @};-@};-
từ @};- và @};-@};-\Rightarrow $(\sqrt{a^2b^2+3a^2}+\sqrt{b^2c^2+3b^2}+\sqrt{c^2c^2+3c^2})^2$ \leq 9+27=36
\Rightarrow$\sqrt{a^2b^2+3a^2}+\sqrt{b^2c^2+3b^2}+\sqrt{c^2c^2+3c^2}$\leq 6 &gt;-&gt;-
Từ &gt;- và &gt;-&gt;- \Rightarrow A \geq $\dfrac{9}{6}$= $\dfrac{3}{2}$
\Rightarrow $MIN_A$=$\dfrac{3}{2}$ tại a=b=c=1

 

[viethoang1999]

86) Cho $\left\{\begin{matrix}a_1;a_2;...;a_n>0 & & \\ a_1+a_2+...+a_n=1 & & \end{matrix}\right.$. Cmr: $\left ( \dfrac{1}{a_1}-1 \right ) \left ( \dfrac{1}{a_2}-1 \right )...\left ( \dfrac{1}{a_n}-1 \right ) \ge (n-1)^n$

87) Cho $x+y+z=0$. Tìm Min $P=\sum \sqrt{3+4^x}$

88) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=3$. Tìm Min $A=\sum \dfrac{a^3}{\sqrt{b^2+3}}$

86)
Có: $1-a_1=a_2+a_3+...+a_n\ge (n-1) \sqrt[n-1]{a_2a_3...a_n}$
Tương tự nhân theo vế ta có:
$(1-a_1)(1-a_2)...(1-a_n)\ge (n-1)^n a_1a_2...a_n$
Suy ra đpcm.
87)
$\sqrt{3+4^x}=\sqrt{1+1+1+4^x}\ge \sqrt{4\sqrt[4]{4^x}}=2\sqrt[8]{4^x}$
\Rightarrow $\sum \sqrt{3+4^x}\ge 2\sum \sqrt[8]{4^x}\ge 2.3\sqrt[3]{\sqrt[8]{4^x.4^y.4^z}}=6\sqrt[24]{4^{x+y+z}}=6$
88)
Cách 2:
$\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{a\sqrt{b^2+3}}{4}\ge a^2$

\Rightarrow $\sum \dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{\sum a\sqrt{b^2+3}}{4}\ge \sum a^2=3$
Áp dụng BCS:
$\left ( \sum a\sqrt{b^2+3} \right )^2\le \sum a^2 \left ( \sum a^2+9 \right ) =36$ \Rightarrow $\sum a\sqrt{b^2+3}\le 6$
Vậy $A\ge 3-\dfrac{6}{4}=\dfrac{3}{2}$
$"="$\Leftrightarrow $a=b=c=1$

 

[viethoang1999]

89) Cho $a;b;c>0$ thỏa: $a+b+c=\dfrac{3}{4}$. Tìm Min $P=\sum \dfrac{1}{\sqrt[3]{a+3b}}$

90) Cho $x;y;z>0$ thỏa: $xyz=1$. Tìm Min $A=\sum \dfrac{x^2(y+z)}{y\sqrt{y}+2z\sqrt{z}}$

89)
Cách 1:
Áp dụng Cauchy 3 số:$\sum \dfrac{1}{\sqrt[3]{a+3b}}\ge 3\sqrt[3]{\dfrac{1}{\sqrt[3]{(a+3b)(b+3c)(c+3a)}}}=\dfrac{3}{\sqrt[9]{(a+3b)(b+3c)(c+3a)}}$
Áp dụng Cauchy 3 số:
$\sqrt[3]{(a+3b)(b+3c)(c+3a)}\le \dfrac{4a+4b+4c}{3}$
$\sqrt[3]{(a+3b)(b+3c)(c+3a)}\le \dfrac{4a+4b+4c}{3}$
\Rightarrow $\sqrt[9]{(a+3b)(b+3c)(c+3a)}\le \sqrt[9]{\dfrac{64(a+b+c)^3}{27}}$
Cách 2:
Áp dụng Cauchy 4 số:
$\dfrac{1}{3\sqrt[3]{a+3b}}+\dfrac{1}{3\sqrt[3]{a+3b}}+\dfrac{1}{3\sqrt[3]{a+3b}}+\dfrac{a+3b}{3}\ge \dfrac{4}{3}$
\Rightarrow $\sum \dfrac{1}{\sqrt[3]{a+3b}}\ge 4-\dfrac{4}{3}\sum a=3$
$"="$\Leftrightarrow $a=b=c=\dfrac{1}{4}$
Cách 3:
$\sum \dfrac{1}{\sqrt[3]{a+3b}}\ge \dfrac{9}{\sum \sqrt[3]{a+3b}}\ge \dfrac{27}{\sum (a+3b+2)}= \dfrac{27}{4(a+b+c)+6}=3$

90)
$x^2(y+z)\ge x^2.2\sqrt{yz}=2x\sqrt{x^2yz}=2x\sqrt{x}$
Đặt $x\sqrt{x}=a;y\sqrt{y}=b;z\sqrt{z}=c$
\Rightarrow $A\ge 2\sum \dfrac{a}{b+2c}=2\sum \dfrac{a^2}{ab+2ac}\ge 2.\dfrac{(\sum a)^2}{3\sum ab}\ge 2.\dfrac{3\sum ab}{3\sum ab}=2$
$"="$\Leftrightarrow $x=y=z=1$

 

[viethoang1999]

91) Cho $\left\{\begin{matrix}a\in [1;2];b\in [4;5];c\in [7;10] & & \\ a+b+c=16 & & \end{matrix}\right.$. Tìm Max $P=abc$

92) Cho $x;y;z>0$ thỏa: $xyz=1$. Tìm Min $P=\sum \dfrac{x^9+y^9}{x^6+x^3y^3+y^6}$

93) Cho $a;b;c>0$ thỏa $abc=1$. Cmr: $\sum \dfrac{1}{a+b+1}\le 1$

94) Cho $a;b;c\ge 0$ thỏa: $ab+bc+ca=3$. Cmr: $\sum \dfrac{1}{1+a^2(b+c)}\le \dfrac{1}{abc}$

95) Cho $a;b;c>0$ thỏa $a^2+b^2+c^2=1$. Cmr: $\sum \dfrac{a}{b^2+c^2}\ge \dfrac{3\sqrt{3}}{2}$