Question

[SA Na]

Giúp vs mọi người ơi

1. a,b,c > 0. C/m: \(\dfrac{c^2}{a+b}+\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}>=\dfrac{a+b+c}{2}\)

2. a,b,c > 0 và a+b+c <= 1. C/m: \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}>=9\)

3. a,b,c là 3 cạnh của một tam giác; \(p=\dfrac{a+b+c}{2}\)
C/m: \(\dfrac{1}{\left(p-a\right)^2}+\dfrac{1}{\left(p-b\right)^2}+\dfrac{1}{\left(p-c\right)^2}>=\dfrac{p}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

4. a,b,c > 0 và (a+c)(b+c)=1
C/m: \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(a+c\right)^2}+\dfrac{1}{\left(b+c\right)^2}>=4\)

Answer 1

[Nue nguyen]

câu 1: \(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)