Question

cho a,b,c la ba so duong .tim gia tri lon nhat cua bieu thuc:

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

Answer 1

Ta có: \(\left(a-1\right)^2\ge0\)

<=> \(a^2-2a+1\ge0\)

<=> \(a^2+1\ge2a\)

=> \(\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)

Tương tự ta cm được: \(\dfrac{b}{b^2+1}\le\dfrac{1}{2}\) ; \(\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)

=> P=\(\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)

dấu bằng sảy ra khi a=b=c=1

vậy P_{MAX =} \(\dfrac{3}{2}\) khi a=b=c=1