Question

[Quach Bich]

Bài 1:Cho \(a,b>0|a^3+b^3=2\).Tìm max \(a+b\)

Bài 2:Cho \(a,b,c>0|abc=1\).Tìm min \(T=\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)

Bài 3:Cho \(a\leq 1; a+\frac{b}{2}\leq 2; a+\frac{b}{2}+\frac{c}{3}\leq 3\)

Tìm min của \(A=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Giup mink với ạThực sự rất cần.Camon nhiều lắm ạ!

Answer 1

[Akai Haruma]

Lời giải:

Bài 1:
Áp dụng BĐT Cô -si ta có:

\(a^3+1+1\geq 3\sqrt[3]{a^3}=3a\)

\(b^3+1+1\geq 3\sqrt[3]{b^3}=3b\)

Cộng theo vế:

\(a^3+b^3+4\geq 3(a+b)\)

\(\Leftrightarrow 6\geq 3(a+b)\Leftrightarrow a+b\leq 2\)

Vậy \((a+b)_{\max}=2\). Dấu bằng xảy ra khi \(a=b=1\)

Bài 2:

Áp dụng BĐT Cô- si ta có:

\(\frac{a^3}{b+c}+\frac{b+c}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{a^3}{8}}=\frac{3}{2}a\)

\(\frac{b^3}{c+a}+\frac{c+a}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{b^3}{8}}=\frac{3}{2}b\)

\(\frac{c^3}{a+b}+\frac{a+b}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{c^3}{8}}=\frac{3}{2}c\)

Cộng theo vế:

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

\(\Leftrightarrow T\geq a+b+c-\frac{3}{2}\)

Theo BĐT Cô-si: \(a+b+c\geq 3\sqrt[3]{abc}=3\)

\(\Rightarrow T\geq 3-\frac{3}{2}=\frac{3}{2}\)

Vậy \(T_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)

Answer 2

[Akai Haruma]

Bài 3:

Điều kiện đề bài tương đương với:

\(a\leq 1; b+2a\leq 4; 2c+3b+6a\leq 18\)

Ta có:

\(A=2\left (\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)+\frac{1}{3}\left(\frac{1}{2a}+\frac{1}{b}\right)+\frac{1}{2a}\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)(6a+3b+2c)\geq (1+1+1)^2\)

\(\Rightarrow \frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\geq \frac{9}{6a+3b+2c}\geq \frac{9}{18}=\frac{1}{2}\) (1)

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

\(\Rightarrow \frac{1}{2a}+\frac{1}{b}\geq \frac{4}{2a+b}\geq \frac{4}{4}=1\) (2)

\(\frac{1}{2a}\geq \frac{1}{2.1}=\frac{1}{2}\) (3)

Từ (1)(2)(3) suy ra \(A\geq 2.\frac{1}{2}+\frac{1}{3}.1+\frac{1}{2}=\frac{11}{6}\)

Dấu bằng xảy ra khi \(a=1; b=2; c=3\)