\(P=\dfrac{a}{\sqrt{1+2bc}}+\dfrac{b}{\sqrt{1+2ca}}+\dfrac{c}{\sqrt{1+2ab}}\)
\(\sqrt{1+2bc}=\sqrt{a^2+b^2+c^2+2bc}=\sqrt{a^2+\left(b+c\right)^2}\ge\sqrt{2\left(ab+ac\right)}\)
Tương tự ta có:
\(\sqrt{1+2ca}\ge\sqrt{2\left(ab+bc\right)}\)
\(\sqrt{1+2ab}\ge\sqrt{2\left(ac+bc\right)}\)
\(P\le\dfrac{a}{\sqrt{2\left(ab+ac\right)}}+\dfrac{b}{\sqrt{2\left(ab+bc\right)}}+\dfrac{c}{\sqrt{2\left(ac+bc\right)}}\)
\(P\le\sqrt{\dfrac{a}{2}.\dfrac{a}{a\left(b+c\right)}}+\sqrt{\dfrac{b}{2}.\dfrac{b}{b\left(a+c\right)}}+\sqrt{\dfrac{c}{2}+\dfrac{c}{c\left(a+b\right)}}\)
\(P\le\sqrt{\dfrac{a}{2}.\dfrac{1}{b+c}}+\sqrt{\dfrac{b}{2}.\dfrac{1}{a+c}}+\sqrt{\dfrac{c}{2}+\dfrac{1}{a+b}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{a}{2}+\dfrac{1}{b+c}+\dfrac{b}{2}+\dfrac{1}{a+c}+\dfrac{c}{2}+\dfrac{1}{a+b}\right)\)
\(P\le\dfrac{1}{2}\left(\dfrac{a+b+c}{2}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\)
______ Bí gòi:v _______
