Question

[Bùi Nguyên Hùng Long]

Có ai bít giải ko vậy

Answer 1

[Đặng Ngọc Quỳnh]

Ta có: \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)

\(\Leftrightarrow b\left(a+b\right)x^2+a\left(a+b\right)y^2\ge ab\left(x+y\right)^2\)

\(\Leftrightarrow abx^2+b^2x^2+a^2y^2+aby^2\ge abx^2+2abxy+aby^2\)

\(\Leftrightarrow b^2x^2+a^2y^2-2abxy\ge0\)

\(\Leftrightarrow\left(bx-ay\right)^2\ge0\) ( bđt luôn đúng)

=> bđt đề bài luôn đúng

Answer 2

[Nguyễn Minh Đăng]

Áp dụng bất đẳng thức Bunyakovsky cho 2 bộ số \(\left(\frac{x}{\sqrt{a}};\frac{y}{\sqrt{b}};\frac{z}{\sqrt{c}}\right)\)  và \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\)

Ta có:\(\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\right)\left(a+b+c\right)\)

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

\(\ge\left(\frac{x}{\sqrt{a}}\cdot\sqrt{a}+\frac{y}{\sqrt{b}}\cdot\sqrt{b}+\frac{z}{\sqrt{c}}\cdot\sqrt{c}\right)^2\)

\(=\left(x+y+z\right)^2\)

\(\Rightarrow\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}\)

Dấu "=" xảy ra khi: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)