\(P^2=\left(\sqrt{bc}.\dfrac{\sqrt{bc}}{\sqrt[4]{a^2+3}}+\sqrt[]{ca}.\dfrac{\sqrt[]{ca}}{\sqrt[4]{b^2+3}}+\sqrt[]{ab}.\dfrac{\sqrt[]{ab}}{\sqrt[4]{c^2+ab}}\right)^2\)
\(P^2\le\left(ab+bc+ca\right)\left(\dfrac{bc}{\sqrt[]{a^2+3}}+\dfrac{ca}{\sqrt[]{b^2+3}}+\dfrac{ab}{\sqrt[]{c^2+ab}}\right)\)
\(P^2\le\dfrac{1}{3}\left(a+b+c\right)^2\left(\dfrac{bc}{\sqrt[]{a^2+3}}+\dfrac{ca}{\sqrt[]{b^2+3}}+\dfrac{ab}{\sqrt[]{c^2+ab}}\right)=3\left(\dfrac{bc}{\sqrt[]{a^2+3}}+\dfrac{ca}{\sqrt[]{b^2+3}}+\dfrac{ab}{\sqrt[]{c^2+ab}}\right)\)
Lại có:
\(ab+bc+ca\le\dfrac{1}{3}\left(a+b+c\right)^2=3\Rightarrow a^2+3\ge a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\dfrac{bc}{\sqrt{a^2+3}}\le\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\)
Tương tự:
\(\dfrac{ca}{\sqrt{b^2+3}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\) ; \(\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Cộng vế:
\(P^2\le\dfrac{3}{2}\left(\dfrac{bc+ca}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{bc+ab}{a+c}\right)=\dfrac{9}{2}\)
\(\Rightarrow P\le\dfrac{3\sqrt{2}}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
