Question

[Nguyễn Hải An]

1, Cho a,b,c là số thực dương và abc =1 . CMR :

\(\dfrac{1}{a^3\left(b+c\right)}\)+ \(\dfrac{1}{b^3\left(c+a\right)}\) + \(\dfrac{1}{c^3\left(a+b\right)}\) \(\ge\)\(\dfrac{3}{2}\)

2, Cho a,b,c là các số thực dương thỏa mãn ab+bc+ca = abc

CM : \(\dfrac{1}{a+2b+3c}\)+ \(\dfrac{1}{2a+3b+c}\) + \(\dfrac{1}{3a+b+2c}\)< \(\dfrac{3}{16}\)

Answer 1

[Akai Haruma]

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)

Tương tự:

\(\frac{1}{b^3(c+a)}+\frac{b(c+a)}{4}\geq \frac{1}{b}=ac\)

\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq \frac{1}{c}=ab\)

Cộng theo vế:

\(\Rightarrow \text{VT}+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)

\(\Rightarrow \text{VT}\geq \frac{ab+bc+ac}{2}\)

Tiếp tục áp dụng AM-GM: \(ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}=3\)

\(\Rightarrow \text{VT}\ge \frac{3}{2}\) (đpcm)

Dấu bằng xảy ra khi $a=b=c=1$

Answer 2

[Akai Haruma]

Lời giải:

Đặt vế trái là $A$

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)(a+b+b+c+c+c)\geq (1+1+1+1+1+1)^2\)

\(\Leftrightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{36}{a+2b+3c}\)

Hoàn toàn TT:

\(\frac{1}{b}+\frac{2}{c}+\frac{3}{a}\geq \frac{36}{b+2c+3a}\)

\(\frac{1}{c}+\frac{2}{a}+\frac{3}{b}\geq \frac{36}{c+2a+3b}\)

Cộng theo vế:

\(\Rightarrow 6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 36A\)

\(\Rightarrow A\leq \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Theo đkđb: \(ab+bc+ac=abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

Do đó: \(A\leq \frac{1}{6}< \frac{3}{16}\) (đpcm)