Tìm GTNN của P=a^3/căn(b^2+3)+b^3/căn(c^2+3)+c^3/căn(a^2+3)

\(P=\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}\)

\(P=\dfrac{a^4}{\sqrt{a^2\left(b^2+3\right)}}+\dfrac{b^4}{\sqrt{b^2\left(c^2+3\right)}}+\dfrac{c^4}{\sqrt{c^2\left(a^2+3\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a^2\left(b^2+3\right)}\le\dfrac{a^2+b^2+3}{2}\\\sqrt{b^2\left(c^2+3\right)}\le\dfrac{b^2+c^2+3}{2}\\\sqrt{c^2\left(a^2+3\right)}\le\dfrac{c^2+a^2+3}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}\le\dfrac{2\left(a^2+b^2+c^2\right)+3}{2}=\dfrac{9}{2}\)

\(\Rightarrow\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\ge\dfrac{2\left(a^2+b^2+c^2\right)^2}{9}=2\)

Vì \(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\)

\(\Rightarrow VT\ge2\)

\(\Leftrightarrow\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}\ge2\)

\(\Leftrightarrow P\ge2\)

Vậy \(P_{min}=2\)