Sorry for not being very descriptive in the title, but here's my problem:(adsbygoogle = window.adsbygoogle || []).push({});

Prove that for all [tex]a,b,c,d \ge 0[/tex], the following inequality is valid:

[tex]\frac{a+b+c+d}{4} \ge \sqrt[4]{abcd}[/tex]

I've tried to make it this way:

[tex](x-y)^2 \ge 0[/tex]

[tex]x^2 + y^2 \ge 2xy[/tex]

So:

[tex][(a+b)-(c+d)]^4 \ge 0[/tex]

[tex][ (a+b)^2 - 2(a+b)(c+d) + (c+d)^2 ] ^2 \ge 0[/tex]

[tex](a+b)^2 + (c+d)^2 \ge 2(a+b)(c+d)[/tex]

[tex](a+b)^2 + 2(a+b)(c+d) + (c+d)^2 \ge 4(a+b)(c+d)[/tex]

[tex]a+b+c+d \ge 2 \sqrt{(a+b)(c+d)}[/tex]

[tex]\frac{a+b+c+d}{2} \ge \sqrt{(a+b)(c+d)}[/tex]

This is the farthest I got, can't figure out how to continue... any tips?

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Another:

Prove, by induction, that the following inequality is true:

[tex]\prod_{n=1} \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}[/tex]

I made this:

[tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{1}{\sqrt{2n+1}} \cdot \frac{2n}{2n+1}[/tex]

[tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)(2n+1)^2}}[/tex]

[tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)^3}}[/tex]

Though I am really new to induction, so I'm not sure if my proof is finished.

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# Inequality proof

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