Question

[Viêt Thanh Nguyễn Hoàng]

Cho a,b,c >0 thỏa mãn : \(a^2+b^2+c^2=abc\\\) .Tìm max của biểu thức :

\(P=\dfrac{a}{a^2+bc}+\dfrac{b}{b^2+ca}+\dfrac{c}{c^2+ab}\)

Answer 1

[Nguyễn Việt Lâm]

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

\(P\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{2}\left(\dfrac{ab+bc+ca}{abc}\right)\le\dfrac{1}{2}\left(\dfrac{a^2+b^2+c^2}{abc}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Answer 2

[Yeutoanhoc]

Áp dụng cosi:

`a^2+bc>=2a\sqrt{bc}`

Hoàn toàn tương tự:

`=>P<=1/2(1/sqrt{ab}+1/sqrt{bc}+1/sqrt{ca})`

Áp dụng cosi:

`1/a+1/b+1/c>=1/sqrt(ab)+1/sqrt(bc)+1/sqrt(ca)`

`=>P<=1/2(1/a+1/b+1/c)`

`=>P<=1/2((ab+bc+ca)/(abc))<=(a^2+b^2+c^2)/(2(abc))=1/2`

Dấu "=" `<=>a=b=c=3`