BĐT cần chứng minh tương đương với:
\(\left(\dfrac{a}{b^2}-\dfrac{2}{b}+\dfrac{1}{a}\right)+\left(\dfrac{b}{a^2}-\dfrac{2}{a}+\dfrac{1}{b}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{16}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2\ge\dfrac{4\left(a-b\right)^2}{ab\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a-b\right)^2}{a^2b^2}\ge\dfrac{4\left(a-b\right)^2}{ab\left(a+b\right)}\).
\(\Leftrightarrow\left(a-b\right)^2\left[\dfrac{a+b}{a^2b^2}-\dfrac{4}{ab\left(a+b\right)}\right]\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{a^2b^2\left(a+b\right)}\ge0\) (luôn đúng).
`a/b^2+b/a^2+16/(a+b)>=5(1/a+1/b)`
`<=>a/b^2-1/b+b^2-1/a+4(4/(a+b)-1/a-1/b)=0`
`<=>(a-b)/b^2+(b-a)/a^2+4((4ab-(a+b)^2)/(ab(a+b)))>=0`
`<=>(a^2(a-b)-b^2(a-b))/(a^2b^2)-(4(a-b)^2)/(ab(a+b))>=0`
`<=>(a-b)^2[(a+b)^2-4ab]>=0`
`<=>(a-b)^2(a^2-2ab+b^2)>=0`
`<=>(a-b)^2(a-b)^2>=0`
`<=>(a-b)^4>=0` luôn đúng.
Dấu "=" xảy ra khi `a=b`
