Ta có:
\(\left(a-b\right)^2\ge0\Leftrightarrow a^2-2ab+b^2\ge0\)\(\Leftrightarrow-a^2-b^2+2ab\le0\Leftrightarrow-a^2-b^2+2ab+2a^2+2b^2\le2a^2+2b^2\)
\(\Leftrightarrow a^2+b^2+2ab\le2a^2+2b^2\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2}\le a^2+b^2\)
\(\Leftrightarrow\dfrac{a+b}{\sqrt{2}}\le\sqrt{a^2+b^2}\)
\(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{2\sqrt{2}}{\sqrt{c}}\ge\dfrac{8}{a+b+c}\)
cm BĐT
Cho a,b,c>0 và thỏa mãn \(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}=\dfrac{7-abc}{\sqrt{2}}\)
Chứng minh rằng \(M=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{3}{2}\)
chứng minh các bát đẳng thức sau
a)Cho a>0 chứng minh rằng \(a+\dfrac{1}{a}\)≥2
b)\(\dfrac{a^2+a+2}{\sqrt{a^2+a+1}}\)≥2
c)\(\sqrt{a+1}-\sqrt{a}< \dfrac{1}{2\sqrt{a}}\)
1 cho a\(\ge0;b\ge0.CMR\)
\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{b}}{2}\)
cho a\(\ge0;b\ge0\). Chứng minh
a)\(a+b\ge2\sqrt{ab}\)
b)\(\sqrt{\dfrac{a+b}{2}}\ge\dfrac{\sqrt{a}+\sqrt{2}}{2}\)
Chứng minh rằng:
a> \(\sqrt{\left(a+c\right)\left(b+d\right)}\ge\sqrt{ab}+\sqrt{cd}\) với a,b,c,d >0
b> \(\dfrac{x^2+5}{\sqrt{x^2+4}}>2\)
cm a+b+\(\dfrac{1}{2}\ge\sqrt{a}+\sqrt{b}\)
cho a,b>1. chứng minh rằng
\(\dfrac{6}{a\sqrt{b-1}+b\sqrt{a-1}}+\sqrt{3ab+4}\ge\dfrac{11}{2}\)
cho a,b,c>0 thỏa mãn abc\(\ge1\)
chứng minh rằng
\(\dfrac{a}{\sqrt{b+\sqrt{ac}}}+\dfrac{b}{\sqrt{c+\sqrt{ab}}}+\dfrac{c}{\sqrt{a+\sqrt{bc}}}\ge\dfrac{3}{\sqrt{2}}\)
