Question

Cho a, b, c > 0; \(ab+bc+ca=3\). Chứng minh rằng: \(P=\dfrac{1}{a^2+2}+\dfrac{1}{b^2+2}+\dfrac{1}{c^2+2}\le1\)

Answer 1

Lời giải:

Ta có:

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

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

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+6}=\frac{(a+b+c)^2}{a^2+b^2+c^2+2(ab+bc+ac)}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)

Do đó: \(2P\leq 3-1=2\Rightarrow P\leq 1\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

Answer 2

@Akai Haruma Giúp e với ạ