Question

[Phượng Hoàng]

Chứng minh rằng:

a, \(4\left(a^3+b^3\right)\ge\left(a+b\right)^3\) với a> b> 0

b, \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

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

Answer 1

[Rimuru tempest]

a) Theo bđt cauchy ta có:

\(a^3+b^3+b^3\ge3\sqrt[3]{a^3.b^6}=3ab^2\)

\(a^3+a^3+b^3\ge3a^2b\)

công vế theo vế ta có \(3\left(a^3+b^3\right)\ge3ab^2+3a^2b\)

\(\Leftrightarrow a^3+b^3+3\left(a^3+b^3\right)\ge a^3+3a^2b+3ab^2+b^3\)

\(\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)

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Answer 2

[Rimuru tempest]

ta luôn có \(\left(a-b\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2\ge2ab\)

\(\Leftrightarrow a^2+b^2+a^2+b^2\ge a^2+2ab+b^2\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow\dfrac{2\left(a^2+b^2\right)}{4}\ge\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow\dfrac{\left(a^2+b^2\right)}{2}\ge\dfrac{\left(a+b\right)^2}{2^2}=\left(\dfrac{a+b}{2}\right)^2\)

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