Toán Cho √x + √y + √z=1 Tim GNNN: x/√y + y/√z + z/√x 05/07/2021 By Elliana Cho √x + √y + √z=1 Tim GNNN: x/√y + y/√z + z/√x
Giải thích các bước giải: Đặt $\sqrt{x}=a, \sqrt y=b,\sqrt z=c$ $\to a+b+c=1$ $\to P=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$ $\to P+1=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c$ $\to P+1=(\dfrac{a^2}{b}+b)+(\dfrac{b^2}{c}+c)+(\dfrac{c^2}{a}+a)$ $\to P+1\ge 2\sqrt{\dfrac{a^2}{b}.b}+2\sqrt{\dfrac{b^2}{c}.c}+2\sqrt{\dfrac{c^2}{a}.a}$ $\to P+1\ge 2a+2b+2c=2\to P\ge 1$ Dấu = xảy ra khi $a=b=c\to x=y=z=\dfrac 19$ Trả lời
Giải thích các bước giải:
Đặt $\sqrt{x}=a, \sqrt y=b,\sqrt z=c$
$\to a+b+c=1$
$\to P=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$
$\to P+1=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+a+b+c$
$\to P+1=(\dfrac{a^2}{b}+b)+(\dfrac{b^2}{c}+c)+(\dfrac{c^2}{a}+a)$
$\to P+1\ge 2\sqrt{\dfrac{a^2}{b}.b}+2\sqrt{\dfrac{b^2}{c}.c}+2\sqrt{\dfrac{c^2}{a}.a}$
$\to P+1\ge 2a+2b+2c=2\to P\ge 1$
Dấu = xảy ra khi $a=b=c\to x=y=z=\dfrac 19$