Question

[Thánh cao su]

\(Cho\) \(a;b;c\ge0\) thỏa mãn \(a+b+c=1\)\(.\)Tìm GTLn của biểu thức:

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

Answer 1

[Akai Haruma]

Lời giải:

Ta có:

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

\(=a^2+1-\frac{b^2(a^2+1)}{b^2+1}+b^2+1-\frac{c^2(b^2+1)}{c^2+1}+c^2+1-\frac{a^2(c^2+1)}{a^2+1}\)

\(=a^2+b^2+c^2+3-\left(\frac{b^2(a^2+1)}{b^2+1}+\frac{c^2(b^2+1)}{c^2+1}+\frac{a^2(c^2+1)}{a^2+1}\right)(*)\)

Vì \(a,b,c\geq 0; a+b+c=1\Rightarrow 0\leq a,b,c\leq 1\)

\(\Rightarrow 0\leq a^2,b^2,c^2\leq 1\)

Do đó:

\(\frac{b^2(a^2+1)}{b^2+1}+\frac{c^2(b^2+1)}{c^2+1}+\frac{a^2(c^2+1)}{a^2+1}\geq \frac{b^2(a^2+1)}{2}+\frac{c^2(b^2+1)}{2}+\frac{a^2(c^2+1)}{2}(**)\)

Từ \((*);(**)\Rightarrow P\leq a^2+b^2+c^2+3-\frac{a^2+b^2+c^2+(a^2b^2+b^2c^2+c^2a^2)}{2}=3+\frac{a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)}{2}\)

Mà: \(a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)=(a+b+c)^2-[2(ab+bc+ac)+(a^2b^2+b^2c^2+c^2a^2]\)

\(=1-[2(ab+bc+ac)+(a^2b^2+b^2c^2+c^2a^2]\leq 1\) do \(a,b,c\geq 0\)

Suy ra \(P\leq 3+\frac{a^2+b^2+c^2-(a^2b^2+b^2c^2+c^2a^2)}{2}\leq 3+\frac{1}{2}=\frac{7}{2}\)

Vậy \(P_{\max}=\frac{7}{2}\Leftrightarrow (a,b,c)=(1,0,0)\) và hoán vị.