Toán cho a,b>0 thoa man a+b=1 chung minh (a+1/b)^2 +(b+1/a)^2>= 25/2 15/07/2021 By Eliza cho a,b>0 thoa man a+b=1 chung minh (a+1/b)^2 +(b+1/a)^2>= 25/2
Giải thích các bước giải: Ta có : $\begin{split}(a+\dfrac{1}b)^2+(b+\dfrac{1}{a})^2&=(a^2+b^2)+(\dfrac{1}{a^2}+\dfrac{1}{b^2})^2+2(\dfrac{a}b+\dfrac{b}a)\\&\ge \dfrac{(a+b)^2}2+\dfrac{1}{2}(\dfrac{1}a+\dfrac{1}b)^2+2.2\sqrt{\dfrac{a}b.\dfrac{b}a}\\&\ge\dfrac{1}2+\dfrac{1}{2}(\dfrac{4}{a+b})^2+4\\&\ge \dfrac{25}2\end{split} $ Dấu = xảy ra khi $a=b=\dfrac{1}2$ Trả lời
Giải thích các bước giải:
Ta có :
$\begin{split}(a+\dfrac{1}b)^2+(b+\dfrac{1}{a})^2&=(a^2+b^2)+(\dfrac{1}{a^2}+\dfrac{1}{b^2})^2+2(\dfrac{a}b+\dfrac{b}a)\\&\ge \dfrac{(a+b)^2}2+\dfrac{1}{2}(\dfrac{1}a+\dfrac{1}b)^2+2.2\sqrt{\dfrac{a}b.\dfrac{b}a}\\&\ge\dfrac{1}2+\dfrac{1}{2}(\dfrac{4}{a+b})^2+4\\&\ge \dfrac{25}2\end{split} $
Dấu = xảy ra khi $a=b=\dfrac{1}2$