Toán a^4 + b^4 ≤ a^6/b^2 +b^6 /a^2 với a,b ≠0 09/07/2021 By Jasmine a^4 + b^4 ≤ a^6/b^2 +b^6 /a^2 với a,b ≠0
Giải thích các bước giải: Ta có :$\dfrac{a^6}{b^2}+a^2b^2\ge 2\sqrt{\dfrac{a^6}{b^2}.a^2b^2}=2a^4$ $\dfrac{b^6}{a^2}+a^2b^2\ge 2\sqrt{\dfrac{b^6}{a^2}.a^2b^2}=2b^4$ $\to \dfrac{a^6}{b^2}+a^2b^2+\dfrac{b^6}{a^2}+a^2b^2\ge 2(a^4+b^4)$ $\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}+2a^2b^2\ge 2(a^4+b^4)$ $\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^4-2a^2b^2+b^4)$ $\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^2-b^2)^2\ge a^4+b^4$ Trả lời
Giải thích các bước giải:
Ta có :
$\dfrac{a^6}{b^2}+a^2b^2\ge 2\sqrt{\dfrac{a^6}{b^2}.a^2b^2}=2a^4$
$\dfrac{b^6}{a^2}+a^2b^2\ge 2\sqrt{\dfrac{b^6}{a^2}.a^2b^2}=2b^4$
$\to \dfrac{a^6}{b^2}+a^2b^2+\dfrac{b^6}{a^2}+a^2b^2\ge 2(a^4+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}+2a^2b^2\ge 2(a^4+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^4-2a^2b^2+b^4)$
$\to \dfrac{a^6}{b^2}+\dfrac{b^6}{a^2}\ge a^4+b^4+(a^2-b^2)^2\ge a^4+b^4$