[tex]\sqrt{\frac{a^2+b^2}{a+b}}+\frac{\sqrt{2ab}}{\sqrt{a+b}}\leq \frac{\sqrt{2(a+b)^2}}{\sqrt{a+b}}=\sqrt{2(a+b)}<=>\sqrt{\frac{a^2+b^2}{a+b}}\leq \sqrt{2(a+b)}-\frac{\sqrt{2ab}}{\sqrt{a+b}}[/tex]
tương tự với 2 cái còn lại nha
[tex]\frac{\sqrt{2ab}}{\sqrt{a+b}}+\frac{\sqrt{2ac}}{\sqrt{a+c}}+\frac{\sqrt{2cb}}{\sqrt{c+b}}\geq 3[/tex]
[tex]\frac{\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}}+\frac{\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{c}}}+\frac{\sqrt{2}}{\sqrt{\frac{1}{c}+\frac{1}{b}}}\geq \frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{c}+\frac{1}{b}}+\sqrt{\frac{1}{a}+\frac{1}{c}}}\geq \frac{9\sqrt{2}}{\sqrt{6\left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )}}[/tex]
mà [tex]ab+bc+ac\leq 3abc<=>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq 3[/tex]
=> [tex]\frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{c}+\frac{1}{b}}+\sqrt{\frac{1}{a}+\frac{1}{c}}}\geq \frac{9\sqrt{2}}{\sqrt{6\left (\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )}}\geq \frac{9\sqrt{2}}{\sqrt{6.3}}=3[/tex]
=> đpcm