Question

a) Cho a, b > 0; Chứng minh rằng: 3(b2 + 2a2) ³ ≥ (b + 2a)2

b) Cho a, b, c > 0 thỏa mãn ab + bc + ca = abc.
chứng minh: \(\dfrac{\sqrt{b^2+2a^2}}{ab}+\dfrac{\sqrt{c^2+2b^2}}{bc}+\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\sqrt{3}\)

Answer 1

b) \(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

\(VT=\dfrac{1}{\sqrt{3}}\left[\dfrac{\sqrt{\left(b^2+2a^2\right)\left(1+2\right)}}{ab}+\dfrac{\sqrt{\left(c^2+2b^2\right)\left(1+2\right)}}{bc}+\dfrac{\sqrt{\left(a^2+2c^2\right)\left(1+2\right)}}{ca}\right]\)

Áp dụng bất đẳng thức Bunhiacopxki:

\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{b+2a}{ab}+\dfrac{c+2b}{bc}+\dfrac{a+2c}{ca}\right)\)

\(=\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}\right)\)

\(=\dfrac{1}{\sqrt{3}}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{\sqrt{3}}.3=\sqrt{3}\left(đpcm\right)\)

- Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Answer 2

Lời giải:
a. Áp dụng BĐT Bunhiacopxky:
$3(b^2+2a^2)=[1^2+(\sqrt{2})^2][b^2+(\sqrt{2}a)^2]\geq (b+2a)^2$
(đpcm) 

b. 

Từ kết quả phần a, suy ra:
$b^2+2a^2\geq \frac{(b+2a)^2}{3}$

$\Rightarrow \sqrt{b^2+2a^2}\geq \frac{b+2a}{\sqrt{3}}$

$\Rightarrow \frac{\sqrt{b^2+2a^2}}{ab}\geq \frac{b+2a}{\sqrt{3}ab}$

Tương tự với các phân thức còn lại:
\(\text{VT}\geq \frac{b+2a}{\sqrt{3}ab}+\frac{c+2b}{\sqrt{3}bc}+\frac{a+2c}{\sqrt{3}ac}=\frac{3(ab+bc+ac)}{\sqrt{3}abc}=\frac{3}{\sqrt{3}}=\sqrt{3}\)

(đpcm) 

Dấu "=" xảy ra khi $a=b=c=3$