Giải:
Ta có:
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}\)
\(\Leftrightarrow P+3=\dfrac{a^3}{\sqrt{1+b^2}}+b^2+\dfrac{b^3}{\sqrt{1+c^2}}+c^2\dfrac{c^3}{\sqrt{1+a^2}}+a^2\)
\(\Leftrightarrow P+\dfrac{6}{4\sqrt{2}}=\dfrac{a^3}{2\sqrt{1+b^2}}+\dfrac{a^2}{2\sqrt{1+b^2}}+\dfrac{1+b^2}{4\sqrt{2}}+\dfrac{b^3}{2\sqrt{1+c^2}}+\dfrac{b^2}{2\sqrt{1+c^2}}+\dfrac{1+c^2}{4\sqrt{2}}+\dfrac{c^3}{2\sqrt{1+a^2}}+\dfrac{c^2}{2\sqrt{1+a^2}}+\dfrac{1+a^2}{4\sqrt{2}}\)
\(\ge3\sqrt[3]{\dfrac{a^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{b^6}{16\sqrt{2}}}+3\sqrt[3]{\dfrac{c^6}{16\sqrt{2}}}\)
\(\Rightarrow P+\dfrac{3}{2\sqrt{2}}\ge\dfrac{3}{2\sqrt[3]{2\sqrt{2}}}\left(a^2+b^2+c^2\right)=\dfrac{9}{2\sqrt[6]{8}}\)
\(\Rightarrow P\ge\dfrac{9}{2\sqrt[6]{2^3}}-\dfrac{3}{2\sqrt{2}}=\dfrac{9}{2\sqrt{2}}-\dfrac{3}{2\sqrt{2}}=\dfrac{3}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐt cauchy-schwarz:(dạng phân thức + đa thức )
\(P=\dfrac{a^3}{\sqrt{1+b^2}}+\dfrac{b^3}{\sqrt{1+c^2}}+\dfrac{c^3}{\sqrt{1+a^2}}=\dfrac{a^4}{a\sqrt{1+b^2}}+\dfrac{b^4}{b\sqrt{1+c^2}}+\dfrac{c^4}{c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\sqrt{1+b^2}+b\sqrt{1+c^2}+c\sqrt{1+a^2}}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{\left(a^2+b^2+c^2\right)\left(3+a^2+b^2+c^2\right)}}=\dfrac{9}{\sqrt{18}}=\dfrac{3}{\sqrt{2}}=\dfrac{3\sqrt{2}}{2}\)
dấu = xảy ra khi a=b=c=1
