Question

[Edogawa Conan]

Cho các số thực dương a;b;c thỏa mãn: \(a\ge c;b\ge c\)

Chứng minh rằng:

\(\sqrt{c.\left(a-c\right)}+\sqrt{c.\left(b-c\right)}\le\sqrt{ab}\)

Answer 1

[Khôi Bùi]

Vì a ; b ; c dương , áp dụng BĐT Cô - si cho các cặp số dương , ta có :

\(\frac{c}{b}+\frac{a-c}{a}\ge2\sqrt{\frac{c\left(a-c\right)}{ab}}\)

\(\frac{c}{a}+\frac{b-c}{b}\ge2\sqrt{\frac{c\left(b-c\right)}{ab}}\)

\(\Rightarrow2\ge2\sqrt{\frac{c\left(a-c\right)}{ab}}+2\sqrt{\frac{c\left(b-c\right)}{ab}}\)

\(\Rightarrow1\ge\frac{\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}}{\sqrt{ab}}\)

\(\Rightarrow\sqrt{ab}\ge\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\)

Dấu " = " xảy ra \(\Leftrightarrow\frac{c}{b}=\frac{a-c}{a};\frac{c}{a}=\frac{b-c}{b}\)

\(\Leftrightarrow\frac{c}{b}+\frac{c}{a}=1\) \(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\)

Vì \(a;b\ge c\Rightarrow a=b=2c\)

Vậy ...

Answer 2

[Nguyễn Việt Lâm]

BĐT cần chứng minh tương đương: \(\sqrt{\frac{c\left(a-c\right)}{ba}}+\sqrt{\frac{c\left(b-c\right)}{ab}}\le1\)

Áp dụng BĐT Cauchy:

\(VT\le\frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}+\frac{c}{a}+\frac{b-c}{b}\right)=\frac{1}{2}\left(\frac{a-c+c}{a}+\frac{c+b-c}{b}\right)=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=2c\)