Question

x^2/a + y^2/b + z^2/c ≥ (x+y+z)^2/a+b+c (a>0,b>0,c>0)

Answer 1

Áp dụng bất đẳng thức Bunhiacopxki cho cặp 3 số ta có:

\(\left[\left(\dfrac{x}{\sqrt{a}}\right)^2+\left(\dfrac{y}{\sqrt{b}}\right)^2+\left(\dfrac{z}{\sqrt{c}}\right)^2\right]\left[\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right]\ge\left[\dfrac{x}{\sqrt{a}}\cdot\sqrt{a}+\dfrac{y}{\sqrt{b}}\cdot\sqrt{b}+\dfrac{z}{\sqrt{c}}\cdot\sqrt{c}\right]^2=\left(x+y+z\right)^2\)

Dấu = xảy ra khi x/a=y/b=z/c

Answer 2

\(\dfrac{x^2}{a}\) + \(\dfrac{y^2}{b}\) + \(\dfrac{z^2}{c}\)≥ \(\dfrac{\left(x+y+z\right)^2}{a+b+c}\)