Toán 8 BDT BUNHIA

[do dinh nam]

[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
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1Cho a;b;c dương tm [tex]a^{2}+b^{2}+c^{2}[/tex][tex]\leq[/tex]abc
cm:[tex]\frac{a}{a^{2}+bc}+\frac{b}{b^{2}+ac}+\frac{c}{c^{2}+ab}\leq \frac{1}{2}[/tex]
2Cho a;b;c>0.
cm:[tex]\frac{\sqrt{a+b}}{\sqrt{a+b+c}}+\frac{\sqrt{b+c}}{\sqrt{a+b+c}}+\frac{\sqrt{a+c}}{\sqrt{a+b+c}}\leq \sqrt{6}[/tex]
3Cho a;b;c>0
cm[tex]\frac{a}{\left ( b+c \right )^{2}}+\frac{b}{\left ( a+c \right )^{2}}+\frac{c}{\left ( a+b \right )^{2}}\geq \frac{9}{4\left ( a+b+c \right )}[/tex]
thanks!!!!

 

[Erwin Schrödinger]

1)
[tex]\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+2ab}\leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2\sqrt{abc}}\leq \frac{\sqrt{3(a+b+c)}}{2\sqrt{abc}}\leq \frac{\sqrt{3.\sqrt{3(a^2+b^2+c^2)}}}{2\sqrt{abc}}\leq \frac{\sqrt{3\sqrt{3}\sqrt{abc}}}{2\sqrt{abc}}=\frac{\sqrt[4]{27}\sqrt[4]{abc}}{2\sqrt{abc}}=\frac{\sqrt[4]{27}}{2\sqrt[4]{abc}}[/tex]
ta lại có : [tex]abc\geq a^2+b^2+c^2\geq 3\sqrt[3]{(abc)^2}<=>(abc)^3\geq 27(abc)^2<=>abc\geq 27[/tex]
=> [tex]\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+2ab}\leq \frac{\sqrt[4]{27}}{2\sqrt[4]{27}}=\frac{1}{2}[/tex]
2)[tex]\frac{\sqrt{b+c}}{\sqrt{a+b+c}}+\frac{\sqrt{a+c}}{\sqrt{a+b+c}}+\frac{\sqrt{b+a}}{\sqrt{a+b+c}}\leq \frac{\sqrt{6(a+b+c)}}{\sqrt{a+b+c}}=\sqrt{6}[/tex]
3) [tex]\frac{\frac{a^2}{4}}{a\frac{b+c}{2}\frac{b+c}{2}}\geq \frac{\frac{a^2}{4}}{\frac{(a+\frac{b+c}{2}+\frac{b+c}{2})^3}{27}}=\frac{27a^2}{4(a+b+c)^3}[/tex]
tương tự với 2 cái còn lại => [tex]\geq \frac{27a^2+27b^2+27c^2}{4(a+b+c)^3}\geq \frac{9(a+b+c)^2}{4(a+b+c)^3}=\frac{9}{4(a+b+c)}[/tex]

 

[Trang Ran Mori]

B1:
[tex]\frac{a}{a^2 + bc}=\frac{a^2}{a^2 +abc}\leq \frac{a^2}{a^3 +a^2 +b^2 +c^2}\leq \frac{1}{4}(\frac{1}{a}+\frac{a^2}{a^2 +b^2 +c^2})[/tex]
Tương tự ta được:
[tex]\frac{a}{a^2 +bc}+\frac{b}{b^2 +ca}+\frac{c}{c^2 +ab}\leq \frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1)[/tex]
Mà [tex]1\geq \frac{a^2 +b^2 +c^2}{abc}\geq \frac{ab +bc+ca}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}[/tex]
Dấu "=" xảy ra khi:a=b=c=3.
B2:
[tex]VT \leq \sqrt{(1^2 +1^2 +1^2)(\frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{c+a}{a+b+c})}=\sqrt{6}[/tex]
Dấu"=" xảy ra khi: a=b=c.
B3:[tex](a+b+c)(\frac{a}{(b+c)^2}+\frac{b}{(c+a)^2}+\frac{c}{(a+b)^2}) =((\sqrt{a})^2+(\sqrt{b})^2+(\sqrt{c})^2)((\frac{\sqrt{a}}{b+c})^2+(\frac{\sqrt{b}}{c+a})^2+(\frac{\sqrt{c}}{a+b})^2) \geq [tex](\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})^2[/tex]
Đặt [tex]P =\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}[/tex]
<=>[tex]P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq \frac{9}{2} => P\geq \frac{3}{2}[/tex]
=> [tex](a+b+c)((\frac{a}{b+c})^2+(\frac{b}{c+a})^2+(\frac{c}{a+b})^2)\geq \frac{9}{4}[/tex]
=>.........
Dấu"=" xảy ra khi: a=b=c.[/tex]