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

H

hien_vuthithanh

95/

AD cauchy
a2.(1a2)2a^2.(1-a^2)^2=12\dfrac{1}{2}.2a2.(1a2).(1a2)2a^2.(1-a^2).(1-a^2) \leq 12\dfrac{1}{2}.(2a2+1a2+1a23)3(\dfrac{2a^2+1-a^2+1-a^2}{3})^3=427\dfrac{4}{27}
\Leftrightarrowa.(1a2) a.(1-a^2) \leq 233\dfrac{2}{3\sqrt{3}} \Leftrightarrow 1a.(1a2)\dfrac{1}{ a.(1-a^2)} \geq 332\dfrac{3\sqrt{3}}{2}\Leftrightarrow a2a.(1a2)\dfrac{a^2}{ a.(1-a^2)} \geq 33a22\dfrac{3\sqrt{3}a^2}{2} \Leftrightarrow a2b2+c2\dfrac{a^2}{b^2+c^2}\geq 33a22\dfrac{3\sqrt{3}a^2}{2}

TT\Rightarrow dpcm

ủng hộ topic của bạn nè!
 
H

hien_vuthithanh

92/

a9+b9x6+x3y3+y6\dfrac{a^9+b^9}{x^6+x^3y^3+y^6}=x3+y32x3y3(x3+y3)x6+x3y3+y6x^3+y^3-\dfrac{2x^3y^3(x^3+y^3)}{x^6+x^3y^3+y^6} \geq x3+y32x3y3(x3+y3)3x9y93x^3+y^3-\dfrac{2x^3y^3(x^3+y^3)}{3\sqrt[3]{x^9y^9}}=13(x3+y3)\dfrac{1}{3}(x^3+y^3)
\Rightarrow a9+b9x6+x3y3+y6\sum \dfrac{a^9+b^9}{x^6+x^3y^3+y^6} \geq 23.x3\dfrac{2}{3}.\sum x^3 \geq 23.3xyz3\dfrac{2}{3}.3\sqrt[3]{xyz}=2
\Rightarrow P min=2 tại x=y=z=1
 
F

forum_

93) Cho a;b;c>0a;b;c>0 thỏa abc=1abc=1. Cmr: 1a+b+11\sum \dfrac{1}{a+b+1}\le 1

Đặt: a=x3a = x^3 ; b=y3b=y^3 ; c=z3c=z^3 . Giả thiết suy ra xyz=1xyz = 1 và yêu cầu chứng minh:

1x3+y3+11\sum \dfrac{1}{x^3+y^3+1}\le 1

Quen quá rồi, hình như có làm ở pic này luôn rồi :D

 
F

forum_

94) Cho a;b;c0a;b;c\ge 0 thỏa: ab+bc+ca=3ab+bc+ca=3. Cmr: 11+a2(b+c)1abc\sum \dfrac{1}{1+a^2(b+c)}\le \dfrac{1}{abc}

Giải:

Từ giả thiết mà Cauchy thì đc abc \leq 1

Suy ra: 1+a2(b+c)1+a^2(b+c) \geq a(ab+bc+ca)=3aa(ab+bc+ca)=3a

Tương tự .... ta có:

11+a2(b+c)\sum \dfrac{1}{1+a^2(b+c)} \leq 13a\sum \dfrac{1}{3a} = 13.(ab+bc+caabc)=1abc\dfrac{1}{3}.(\dfrac{ab+bc+ca}{abc}) = \dfrac{1}{abc}

Đẳng thức chứng minh xong
 
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V

viethoang1999

91) Cho {a[1;2];b[4;5];c[7;10]a+b+c=16\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=abcP=abc
Làm bài trích đề cho tiện theo dõi nhé!
91)
Có: a<b<ca<b<c
Dự đoán dấu "=""=" xảy ra khi: αa=βb=c\alpha a=\beta b=c với α>β>1\alpha >\beta >1
Xét: α.β.P=αa.βb.c(αa+βb+c3)3=[(α1)a+(β1)b+163]3[(α1).2+(β1).5+163]3\alpha. \beta .P=\alpha a.\beta b.c\le \left ( \dfrac{\alpha a+\beta b+c}{3} \right ) ^3=\left [ \dfrac{(\alpha -1)a+(\beta -1)b+16}{3} \right ] ^3\le \left [ \dfrac{(\alpha -1).2+(\beta -1).5+16}{3} \right ]^3
"=""=" \Leftrightarrow {a=2b=5c=9αa=βb=c\left\{\begin{matrix}a=2 & & \\ b=5 & & \\c=9 \\\alpha a=\beta b=c \end{matrix}\right.
\Leftrightarrow {α=92β=95\left\{\begin{matrix}\alpha =\dfrac{9}{2} & & \\ \beta =\dfrac{9}{5} & & \end{matrix}\right.
Xét: 92a.95b.c\dfrac{9}{2}a.\dfrac{9}{5}b.c
Làm tương tự như bên trên ta có Max.

 
V

viethoang1999

96) Cho x;y;z>0x;y;z>0. Cmr: 2xx3+y21x2\sum\dfrac{2\sqrt{x}}{x^3+y^2}\le \sum \dfrac{1}{x^2}

97) Cho a;b;c>0a;b;c>0. Cmr: 1a2+bca+b+c2abc\sum \dfrac{1}{a^2+bc}\le \dfrac{a+b+c}{2abc}

98) Cho {x;y;z>0x+y+z=1\left\{\begin{matrix}x;y;z>0 & & \\ x+y+z=1 & & \end{matrix}\right.. Cmr: x2+xy+y23\sum \sqrt{x^2+xy+y^2}\ge \sqrt{3}

99) Cho {x;y;z>0x+y+z=1\left\{\begin{matrix}x;y;z>0 & & \\ x+y+z=1 & & \end{matrix}\right.. Cmr: 2x2+xy+y25\sum \sqrt{2x^2+xy+y^2}\ge \sqrt{5}

100) Cho a;b>0a;b>0. Tìm Max Q=a3+b3Q=a^3+b^3 biết a+b=a2ab+b2a+b=a^2-ab+b^2

101) Cho a;b;c>0a;b;c>0 thỏa a2+b2+c2=1a^2+b^2+c^2=1. Cmr: 1a2+b2a32abc+3\sum \dfrac{1}{a^2+b^2}\le \dfrac{\sum a^3}{2abc}+3
 
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E

eye_smile

96) Cho x;y;z>0x;y;z>0. Cmr: 2xx3+y21x2\sum\dfrac{2\sqrt{x}}{x^3+y^2}\le \sum \dfrac{1}{x^2}

Có: x3+y22xyxx^3+y^2 \ge 2xy\sqrt{x}

\Rightarrow 2xx3+y22x2xyx=1xy12(1x2+1y2)\dfrac{2\sqrt{x}}{x^3+y^2} \le \dfrac{2\sqrt{x}}{2xy\sqrt{x}}=\dfrac{1}{xy} \le \dfrac{1}{2}(\dfrac{1}{x^2}+\dfrac{1}{y^2})

Tương tự, cộng theo vế \Rightarrow đpcm
 
E

eye_smile


Có: a2+bc2abca^2+bc \ge 2a\sqrt{bc}

\Rightarrow 1a2+bc12abc=bc2abcb+c4abc\dfrac{1}{a^2+bc} \le \dfrac{1}{2a\sqrt{bc}}=\dfrac{\sqrt{bc}}{2a\sqrt{bc}} \le \dfrac{b+c}{4abc}

Tương tự, cộng theo vế \Rightarrow đpcm
 
E

eye_smile

98) Cho {x;y;z>0x+y+z=1\left\{\begin{matrix}x;y;z>0 & & \\ x+y+z=1 & & \end{matrix}\right.. Cmr: x2+xy+y23\sum \sqrt{x^2+xy+y^2}\ge \sqrt{3}

x2+xy+y2=(x+12y)2+(32y)2(x+12y+y+12z+z+12x)2+34(x+y+z)2=3\sum \sqrt{x^2+xy+y^2} =\sum \sqrt{(x+\dfrac{1}{2}y)^2+(\dfrac{\sqrt{3}}{2}y)^2} \ge \sqrt{(x+\dfrac{1}{2}y+y+\dfrac{1}{2}z+z+\dfrac{1}{2}x)^2+\dfrac{3}{4}(x+y+z)^2} =\sqrt{3}

99) Cho {x;y;z>0x+y+z=1\left\{\begin{matrix}x;y;z>0 & & \\ x+y+z=1 & & \end{matrix}\right.. Cmr: 2x2+xy+y25\sum \sqrt{2x^2+xy+y^2}\ge \sqrt{5}

2x2+xy+y2=2(x+14y)2+78y22(x+y+z+14x+14y+14z)2+78(x+y+z)2=2\sum \sqrt{2x^2+xy+y^2} =\sum \sqrt{2(x+\dfrac{1}{4}y)^2+\dfrac{7}{8}y^2} \ge \sqrt{2(x+y+z+\dfrac{1}{4}x+\dfrac{1}{4}y+\dfrac{1}{4}z)^2+\dfrac{7}{8}(x+y+z)^2}=2

Liệu đề câu này có nhầm không nhỉ?
 
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T

tranvanhung7997

98 (99 tương tự) x2+xy+y2=(x+y2)2+(3y2)2((x+y2))2+(3y2)2=3\sum \sqrt{x^2+xy+y^2} = \sum \sqrt{(x+\dfrac{y}{2})^2 + (\dfrac{\sqrt{3}y}{2})^2} \ge \sqrt{(\sum (x+\dfrac{y}{2}))^2 + (\sum \dfrac{\sqrt{3}y}{2})^2} = \sqrt{3} (Mincôpski)
=> đpcm
 
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E

eye_smile

100) Cho a;b>0a;b>0. Tìm Max Q=a3+b3Q=a^3+b^3 biết a+b=a2ab+b2a+b=a^2-ab+b^2

Ta có: Q=(a+b)(a2ab+b2)=(a+b)2Q=(a+b)(a^2-ab+b^2)=(a+b)^2

Lại có: a+b=a2ab+b2=(a+b)23ab14(a+b)2a+b=a^2-ab+b^2=(a+b)^2-3ab \ge \dfrac{1}{4}(a+b)^2

\Rightarrow a+b4a+b \le 4

\Rightarrow Q16Q \le 16

Dấu "=" xảy ra \Leftrightarrow a=b=2a=b=2
 
E

eye_smile


Ta có: 1a2+b2=1+c2a2+b21+c22ab=1+c32abc\dfrac{1}{a^2+b^2}=1+\dfrac{c^2}{a^2+b^2} \le 1+\dfrac{c^2}{2ab}=1+\dfrac{c^3}{2abc}

Tương tự, cộng theo vế \Rightarrow đpcm
 
V

viethoang1999


99) Cho {x;y;z>0x+y+z=1\left\{\begin{matrix}x;y;z>0 & & \\ x+y+z=1 & & \end{matrix}\right.. Cmr: 2x2+xy+y25\sum \sqrt{2x^2+xy+y^2}\ge \sqrt{5}

2x2+xy+y2=x2+34(x+y)2+14(xy)2\sum \sqrt{2x^2+xy+y^2}=\sum \sqrt{x^2+\dfrac{3}{4}(x+y)^2+\dfrac{1}{4}(x-y)^2}
x2+3(x+y)24=124x2+3(x+y)2\ge \sum \sqrt{x^2+\dfrac{3(x+y)^2}{4}}=\dfrac{1}{2}\sum \sqrt{4x^2+3(x+y)^2}
=12(2x)2+(x+y)2+(x+y)2+(x+y)2122x+x+y+x+y+x+y4=\dfrac{1}{2}\sum \sqrt{(2x)^2+(x+y)^2+(x+y)^2+(x+y)^2}\ge \dfrac{1}{2}\sum \dfrac{2x+x+y+x+y+x+y}{4}
=12.5x+3y4=x=1=\dfrac{1}{2}.\sum \dfrac{5x+3y}{4}=\sum x=1
 
V

viethoang1999

102) Cho a;b;c>0a;b;c>0 thỏa abc=2abc=2. Cmr: a3ab+c\sum a^3\ge \sum a\sqrt{b+c}

103) Cho a;b>0a;b>0 thỏa a2+b2=1a^2+b^2=1. Tìm Min A=1a+1b(abba)2A=\dfrac{1}{a}+\dfrac{1}{b}-\left ( \sqrt{\dfrac{a}{b}}-\sqrt{\dfrac{b}{a}}\right )^2

104) Cho 2 dãy an;bn>0{a_n};{b_n}>0 thoả: an+1=an+1ana_{n+1}=a_n+\dfrac{1}{a_n} ; bn+1=bn+1bnb_{n+1}=b_n+\dfrac{1}{b_n}.
Cmr: C25=a25+b25>102C_{25}=a_{25}+b_{25}>10\sqrt{2}

105) Cho a;b>0a;b>0. Cmr:
ab+(ab)2(3a+b)(a+3b)8(a+b)(a2+6ab+b2)a+b2\sqrt{ab}+\dfrac{(a-b)^2(3a+b)(a+3b)}{8(a+b)(a^2+6ab+b^2)}\le \dfrac{a+b}{2}

106) Cho a;b;c>0a;b;c>0 thỏa: a2+b2+c2=4abca^2+b^2+c^2=4\sqrt{abc}
Cmr: a+b+c>2abca+b+c>2\sqrt{abc}
 
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