2 replies to this topic

[misakichan]

Cho a, b, c, m, n, p>0

CMR: $\sqrt[3]{(a+m)(b+n)(c+p)}\geq \sqrt[3]{abc}+\sqrt[3]{mnp}$

[Baoriven]

Chỉ cần dùng Holder.

Ta có: $(a+m)(b+n)(c+p)=(a+\frac{m}{2}+\frac{m}{2})(b+\frac{n}{2}+\frac{n}{2})(c+\frac{p}{2}+\frac{p}{2})$

$\geq (\sqrt[3]{abc}+\sqrt[3]{\frac{mnp}{8}}.2)^3=(\sqrt[3]{abc}+\sqrt[3]{mnp})^3$

Vậy ta có đpcm.

[tienduc]

Cho a, b, c, m, n, p>0

CMR: $\sqrt[3]{(a+m)(b+n)(c+p)}\geq \sqrt[3]{abc}+\sqrt[3]{mnp}$

Bất đẳng thức cần cm tương đương với 

$\sqrt[3]{\frac{a}{a+m}.\frac{b}{b+n}.\frac{c}{c+p}}+\sqrt[3]{\frac{m}{a+m}.\frac{n}{b+n}.\frac{p}{c+p}}\leq 1$

Áp dụng BĐT $AM-GM$ ta có 

$\sqrt[3]{\frac{a}{a+m}.\frac{b}{b+n}.\frac{c}{c+p}}\leq \frac{\frac{a}{a+m}+\frac{b}{b+n}+\frac{c}{c+n}}{3}$

$\sqrt[3]{\frac{m}{a+m}.\frac{n}{b+n}.\frac{p}{c+p}}\leq \frac{\frac{m}{a+m}+\frac{n}{b+n}+\frac{p}{c+p}}{3}$

Cộng vế có đpcm

Edited by tienduc, 05-01-2017 - 21:04.