Question

Cho a,b,c > 0 và ab+bc+ca = 1. Tìm: \(MaxP=\dfrac{a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+c^2}}+\dfrac{c}{\sqrt{1+c^2}}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{a}{\sqrt{1+a^2}}=\dfrac{a}{\sqrt{ab+bc+ca+a^2}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\)

\(\le\dfrac{1}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\). Thiết lập 2 BĐT tương tự:

\(\dfrac{b}{\sqrt{1+b^2}}\le\dfrac{1}{4}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{1+c^2}}\le\dfrac{1}{4}\left(\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(P\le\dfrac{1}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)