Inequalities!

Can you prove this? $(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \ge (a+b+c)^3 .$

#NumberTheory #Inequalities #OlympiadMath #OlympaidPreparation

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

We can solve it using Holder's inequality..this problem appeared in USAMO 2004

Ayush Garg - 5 years, 10 months ago

can you write the solution to this problem?

Gokul Kumar - 5 years, 10 months ago

http://www.artofproblemsolving.com/wiki/index.php/2004USAMOProblems/Problem_5

Edit: if the link doesn't work, just search "USAMO 2004 problem 5" on Google and you'll find an AoPS link with full solutions.

mathh mathh - 5 years, 9 months ago

First use rearrangement inequality and then Holder

kushal padole - 5 years, 5 months ago

Yeah thanks

Gokul Kumar - 5 years, 4 months ago

You can find some full solutions on AoPS after searching "USAMO 2004 Problem 5" on Google.

mathh mathh - 5 years, 4 months ago

Ya I saw that long back. I just felt like thanks for writing his suggestion.

Gokul Kumar - 5 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$