đè sai r ,,,,thử a=b=c=3 xem. ok??
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Cho a,b,c>0. Cmr:
\(\frac{a}{\sqrt{ab+b^2}}+\frac{b}{\sqrt{bc+b^2}}+\frac{c}{\sqrt{ac+c^2}}\ge\frac{3\sqrt{2}}{2}\)
cho a,b,c>0;ab+bc+ac\(\le\)3abc
cmr\(\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{a^2+c^2}{a+c}}+3\le\sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}\right)\)
cho a,b,c>0 thỏa mãn \(a^2+b^2+c^2=1\).CMR
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}+\dfrac{\sqrt{bc+2a^2}}{\sqrt{1+bc-a^2}}+\dfrac{\sqrt{ca+2b^2}}{\sqrt{1+ca-b^2}}\ge2+ab+bc+ca\)
6 tháng 9 2021 lúc 18:32
cho a,b,c>0.cmr
\(\sqrt{a^2+2b^2+ab}+\sqrt{b^2+2c^2+bc}+\sqrt{c^2+2a^2+ac}\ge2\left(a+b+c\right)\)
CMR:\(\sqrt{\frac{a^2+2b^2}{a^2+ab+bc}}+ \sqrt{\frac{b^2+2c^2}{b^2+bc+ac}}+\sqrt{\frac{c^2+2a^2}{c^2+ab+ac}} \geq 3\)
Cho a, b, c >0 tm \(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=6\)
CMR \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}\ge3\)
Cho a,b,c >0 thỏa mãn a+b+c=1. CMR:
\(P=\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ac}{b+ac}}+\sqrt{\frac{ab}{c+ab}}\le\frac{3}{2}\)
CMR :
\(\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{a+c}{ac}}+\sqrt{\frac{b+c}{bc}}\)
cho a, b, c>0. Tìm max:
P=\(\frac{\sqrt{bc}}{a+2\sqrt{bc}}+\frac{\sqrt{ac}}{b+2\sqrt{ac}}+\frac{\sqrt{ab}}{c+2\sqrt{ab}}\)