`a;b>0;a^2+b^2=2`
Áp dụng BĐT Côsi ta có:
+)`\qquad a^2+b^2\ge 2\sqrt{a^2b^2}=2ab`
`<=>2\ge 2ab<=>ab\le 1<=>\sqrt{ab}\le 1`
`<=>1/{\sqrt{ab}}\ge 1`
+) `(a/b + b/a)\ge 2\sqrt{a/b . b/a}=2`
+) `(a/{b^2}+b/{a^2})\ge 2\sqrt{a/{b^2} . b/{a^2}}=2/{\sqrt{ab}}`
`=>(a/b + b/a)(a/{b^2}+b/{a^2})\ge 2. 2/{\sqrt{ab}}=4/{\sqrt{ab}}=4`
Vậy `(a/b + b/a)(a/{b^2}+b/{a^2})\ge 4`.
Dấu "$=$" xảy ra khi $a=b=1$
