Bài 1:Cho a,b,c,d là các số dương. Chứng minh rằng :
\(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}+\frac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\ge\frac{a+b+c+d}{4}\)
Bài 2:Cho \(a>0,b>0,c>0\).\(CM:\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài 3: a) Cho x,y,>0. CMR:\(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\)
b) Chứng minh rằng\(\Sigma\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
dễ cm \(\frac{9\left(a+b\right)\left(b+c\right)\left(c+a\right)}{4\left(ab+bc+ca\right)}\ge2\left(a+b+c\right)\)
\(2\sqrt{a^2-ab+b^2}=2\sqrt{\left(\frac{a^2}{b}-a+b\right)b}\le a^2-a+2b\)
từ đó bđt cần cm <=> \(a+b+c\ge ab+bc+ca\)
lại có \(ab+bc+ca+abc\le4\)
\(\Leftrightarrow\left(a+2\right)\left(b+2\right)\left(c+2\right)\le\left(a+2\right)\left(b+2\right)+...\)
\(\Leftrightarrow\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\ge1\)
\(\Leftrightarrow\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\le\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\le1\)
\(\Rightarrow a+b+c\ge ab+bc+ca\)
=>Q.E.D
đặt \(P=\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\)
\(\Rightarrow P-3=\frac{ab}{1-ab}+\frac{bc}{1-bc}+\frac{ca}{1-ca}\le\frac{ab}{1-\frac{a^2+b^2}{2}}+\frac{bc}{1-\frac{b^2+c^2}{2}}+\frac{ca}{1-\frac{c^2+a^2}{2}}\)
\(\le\frac{1}{2}.\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}+\frac{1}{2}.\frac{\left(b+c\right)^2}{\left(a^2+b^2\right)+\left(c^2+a^2\right)}+\frac{1}{2}.\frac{\left(c+a\right)^2}{\left(b^2+c^2\right)+\left(b^2+a^2\right)}\)
\(\le\frac{1}{2}.\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+a^2}+\frac{c^2}{b^2+c^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{2}\)
\(\Rightarrow P-3\le\frac{3}{2}\Rightarrow P\le\frac{9}{2}\)
các bạn làm được ý nào thì làm ý đó nha
1. Cho a,b,c là độ dài 3 cạnh tam giác. Chứng minh:
a) \(\frac{1}{\left(a+b-c\right)^2}+\frac{1}{\left(a-b+c\right)^2}+\frac{1}{\left(b+c-a\right)^2}\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
b) \(\frac{1}{\left(a+b-c\right)^3}+\frac{1}{\left(a-b+c\right)^3}+\frac{1}{\left(b+c-a\right)^3}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\)
c) \(\frac{1}{\left(a+b-c\right)^{200}}+\frac{1}{\left(a-b+c\right)^{200}}+\frac{1}{\left(b+c-a\right)^{200}}\ge\frac{1}{a^{200}}+\frac{1}{b^{200}}+\frac{1}{c^{200}}\)
d) \(\frac{1}{8}\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\sqrt{abc\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}\)
e) \(a+b+c< \sqrt{a\left(b+c\right)}+\sqrt{b\left(a+c\right)}+\sqrt{c\left(a+b\right)}\)
f) \(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}< \sqrt{6}\)
g) \(\sqrt{-a+b+c}+\sqrt{a-b+c}+\sqrt{a+b-c}\le\sqrt{3\left(a+b+c\right)}\)
\(P=\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}+\frac{b}{\sqrt{\left(c+1\right)\left(c^2-c+1\right)}}+\frac{c}{\sqrt{\left(a+1\right)\left(a^2-a+1\right)}}\)
\(\ge\frac{2a}{b^2+2}+\frac{2b}{c^2+2}+\frac{2c}{a^2+2}=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+2}+\frac{bc^2}{c^2+2}+\frac{ca^2}{a^2+2}\right)\)
\(=6-\left(\frac{2ab^2}{b^2+4+b^2}+\frac{2bc^2}{c^2+4+c^2}+\frac{2ca^2}{a^2+4+a^2}\right)\ge6-\left(\frac{2ab}{b+4}+\frac{2bc}{c+4}+\frac{2ca}{a+4}\right)\)
\(=6-\left(2a+2b+2c-\frac{8a}{b+4}-\frac{8b}{c+4}-\frac{8c}{a+4}\right)\)
\(=\frac{8a}{b+4}+\frac{8b}{c+4}+\frac{8c}{a+4}-6=\frac{8a^2}{ab+4a}+\frac{8b^2}{bc+4b}+\frac{8c^2}{ca+4c}-6\)
\(\ge\frac{8\left(a+b+c\right)^2}{\left(ab+bc+ca\right)+4\left(a+b+c\right)}-6\ge\frac{288}{\frac{\left(a+b+c\right)^2}{3}+24}-6=2\)
1) Cho a,b,c>0 tm a+b+c=3. Cmr \(\frac{1}{2+a^2+b^2}+\frac{1}{2+b^2+c^2}+\frac{1}{2+c^2+a^2}\le\frac{3}{4}\)
2) Cho a,b,c>0 tm a^2+b^2+c^2 bé hơn hoặc bằng abc. Cmr \(\frac{a}{a^2+bc}+\frac{b}{b^2+ca}+\frac{c}{c^2+ab}\le\frac{1}{2}\)
3) Cho a,b,c>0 tm a+b+c<=3. Cmr \(\frac{ab}{\sqrt{3+c}}+\frac{bc}{\sqrt{3+a}}+\frac{ca}{\sqrt{3+b}}\le\frac{3}{2}\)
4) Cho a,b,c>0 tm a+b+c=2. Cmr \(\frac{a}{\sqrt{4a+3bc}}+\frac{b}{\sqrt{4b+3ca}}+\frac{c}{\sqrt{4c+3ab}}\le1\)
5) Cho a,b,c>0. Cmr \(\sqrt{\frac{a^3}{5a^2+\left(b+c\right)^2}}+\sqrt{\frac{b^3}{5b^2+\left(c+a\right)^2}}+\sqrt{\frac{c^3}{5c^2+\left(a+b\right)^2}}\le\sqrt{\frac{a+b+c}{3}}\)
6) Cho a,b,c>0. Cmr \(\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}+\frac{b^2}{\left(2b+a\right)\left(2b+c\right)}+\frac{c^2}{\left(2c+a\right)\left(2c+b\right)}\le\frac{1}{3}\)
Giúp mình với nhé các bạn
dùng AM-GM nha
a) cm \(\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)với \(c>0;a,b\ge c\)
b) \(\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+d\right)\left(b+c\right)}\)với a,b,c,d>0
c) cho a,b,c,d>0
cm \(\sqrt{\frac{a}{b+c+d}}+\sqrt{\frac{b}{a+c+d}}+\sqrt{\frac{c}{a+b+d}}+\sqrt{\frac{d}{a+b+c}}>2\)
Cho các số thực dương a,b,c thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\). CMR:
\(\frac{a+b}{\sqrt{ab+c}}+\frac{b+c}{\sqrt{bc+a}}+\frac{c+a}{\sqrt{ca+b}}\ge3\sqrt[6]{abc}\)
Giải:
\(GT\Leftrightarrow ab+bc+ca\ge abc\)
\(\Rightarrow ab\le\frac{ab+bc+ca}{c}\)
\(\Rightarrow\frac{a+b}{\sqrt{ab+c}}\ge\frac{a+b}{\sqrt{\frac{ab+bc+ca}{c}+c}}=\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
Tương tự rồi cộng lại: \(VT\ge\frac{\left(a+b\right)\sqrt{c}}{\sqrt{\left(c+a\right)\left(c+b\right)}}+\frac{\left(b+c\right)\sqrt{a}}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{\left(c+a\right)\sqrt{c}}{\sqrt{\left(b+a\right)\left(b+c\right)}}\)\(\ge3\sqrt[3]{\sqrt{abc}}=3\sqrt[6]{abc}\)
Lần sau mấy bạn hỏi bài thì đăng lên nhé!