Tìm GTLN của P=a/căn(a+a^2)+b/căn(1+b^2)+c/căn(1+c^2)

☘ Ta có:

\(P=\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}\)

\(=\dfrac{a}{\sqrt{ab+ac+ca+a^2}}+\dfrac{b}{\sqrt{ab+ac+ca+b^2}}+\dfrac{c}{\sqrt{ab+ac+ca+c^2}}\)

\(=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

☘ Áp dụng bất đẳng thức AM - GM

\(\Rightarrow\dfrac{1}{\sqrt{a+b}}\times\dfrac{1}{\sqrt{a+c}}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{2}\)

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

☘ Tương tự, ta cũng có:

\(\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}\le\dfrac{b}{2\left(a+b\right)}+\dfrac{b}{2\left(b+c\right)}\)

\(\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{c}{2\left(a+c\right)}+\dfrac{c}{2\left(b+c\right)}\)

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

☘ Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

⚠ Source: https://hoc24.vn/hoi-dap/question/237527.html