\(\dfrac{\dfrac{b+c}{a}+\dfrac{a}{a}}{2}>=\sqrt{\dfrac{b+c}{a}\cdot\dfrac{a}{a}}\)
=>\(\dfrac{a+b+c}{2a}>=\sqrt{\dfrac{b+c}{a}}\)
=>\(\sqrt{\dfrac{a}{b+c}}>=\dfrac{2a}{a+b+c}\)
Tương tự, ta có: \(\sqrt{\dfrac{b}{a+c}}>=\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}>=\dfrac{2c}{a+b+}\)
=>A>=2
Cho a, b, c > 0. CMR :
\(\dfrac{\sqrt{a^2+b^2}}{c}+\dfrac{\sqrt{b^2+c^2}}{a}+\dfrac{\sqrt{a^2+c^2}}{b}\ge2\left(\dfrac{a}{\sqrt{b^2+c^2}}+\dfrac{b}{\sqrt{a^2+c^2}}+\dfrac{c}{\sqrt{a^2+b^2}}\right)\)
Cho a,b∈Z, c≠0 và \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
CMR: \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}>=\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{c^2+a^2}{2}}\left(a,b,c>0\right)\)
Cho a,b,c>0.CMR \(\sqrt{1+\dfrac{16a}{b+c}}+\sqrt{1+\dfrac{16b}{a+c}}+\sqrt{1+\dfrac{16c}{a+b}}\ge9\)
Cho a,b,c>0 thỏa mãn ab+bc+ca=1. CMR:
\(\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^3\le\dfrac{3}{2}\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
Cho các số thực dương a+b+c=\(\sqrt{a}+\sqrt{b}+\sqrt{c}=2\\\).CMR
\(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}=\dfrac{2}{\sqrt{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
cho a,b,c>0 cm \(\dfrac{a+b}{\sqrt{c}}\) +\(\dfrac{b+c}{\sqrt{a}}\) +\(\dfrac{c+a}{\sqrt{b}}\)≥2(\(\sqrt{a}+\sqrt{b}+\sqrt{c}\))
cho a,b,c>0 thỏa mãn \(a^2+b^2+c^2=1\).CMR
\(\dfrac{\sqrt{ab+2c^2}}{\sqrt{1+ab-c^2}}+\dfrac{\sqrt{bc+2a^2}}{\sqrt{1+bc-a^2}}+\dfrac{\sqrt{ca+2b^2}}{\sqrt{1+ca-b^2}}\ge2+ab+bc+ca\)
Cho a,b,c > 0 thỏa mãn \(a\sqrt{\dfrac{b}{c}}+b\sqrt{\dfrac{c}{a}}+c\sqrt{\dfrac{a}{b}}=3\). Chứng minh rằng:
\(N=\dfrac{a^4}{b^2}+\dfrac{b^4}{c^2}+\dfrac{c^4}{a^2}\ge3\)
