a. Xét hiệu: \(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}\)
=\(\dfrac{b\left(a+b\right)+a\left(a+b\right)-4ab}{ab\left(a+b\right)}\)
\(=\dfrac{a^2-2ab+b^2}{ab\left(a+b\right)}=\dfrac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\)
Vì a,b>0
Xảy ra đẳng thức khi và chỉ khi a=b
a) Ta có: \(\left(a-b\right)^2\ge0\left(1\right)\forall a,b\)
( Dấu = xày ra khi và chỉ khi a=b)
Cộng 4ab vào 2 vế, ta có:
\(\left(a-b\right)^2+4ab\ge4ab\Leftrightarrow\left(a+b\right)^2\ge4ab\)
Chia 2 vế cho ab(a+b)>0, ta có:
\(\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
b) Ta có:
\(2p=a+b+c\)
\(p-a=\dfrac{a+b+c}{2}-a=\dfrac{b+c-a}{2}>0\) vì b+c>a
Tương tự: \(p-b>0,p-c>0\)
Áp dụng BĐT: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)cho từng cặp số p-a, p-b; p-b,p-c;p-c,p-a
Ta có:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{\left(p-a\right)+\left(p-b\right)}=\dfrac{4}{2p-\left(a+b\right)}=\dfrac{4}{c}\left(1\right)\)
Tương tự:
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\left(2\right)\)
\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\left(3\right)\)
Cộng các BĐT cùng chiều (1), (2), (3) vế theo vế, ta có:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Do đó: \(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
