Would like help getting started, unfamiliar with proving inequalities in general.
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2This is known as Weighted AM-GM Inequality. Check out proofs here and here. – TenaliRaman Apr 07 '15 at 04:39
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this is a simple and pretty form of a very widely-used result, so it is worth looking at it from many different points of view. here is one -
suppose $b \in (0,1]$. let $s$ denote a real variable on $[1,\infty)$. and consider the function: $$ f(s) = 1 + b(s-1) - s^b $$ clearly $f(1)=0$. and we have $$ f'(s) = b(1-s^{b-1}) \ge 0 $$ with the inequality strict if $s \gt 1$.
this means that $f \ge 0 $, which we may write as: $$ s^b \le 1- b + bs \tag{1} $$ suppose $y \ge x \gt 0$ and set $s=\frac{y}{x}$. then (1) becomes: $$ \left(\frac{y}{x} \right)^b \le (1-b) + b\left(\frac{y}{x} \right) $$ multiplying both sides by $x$ now gives: $$ x^{1-b}y^b \le (1-b)x + by $$ writing $a$ for $1-b$ gives the result required
David Holden
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Consider, $f(x)=\log(x)$. Then $f''(x)=-\frac{1}{x^2}<0$. So, $f$ is concave function. Since $a+b=1$ so, using Jensen's inequality ,
$$f(ax+by)\ge af(x)+bf(y).$$
$$\implies \log(ax+by)\ge a\log x+b\log y$$
$$\implies ax+by\ge x^ay^b.$$
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Using Jensen's inequality:
$f(\frac{\sum_i a_i x_i}{\sum_i a_i}) \geq \frac{\sum_i a_i f(x_i)}{\sum_i a_i}$ for a concave function $f$.
$f=log(x)$ is concave.
Thus,
$log(\frac{\sum_i a_i x_i}{\sum_i a_i}) \geq \frac{\sum_i a_i log(x_i)}{\sum_i a_i}$ $\implies$ $log(\frac{\sum_i a_i x_i}{\sum_i a_i}) \geq log((\Pi x_i^{a_i})^{\frac{1}{\sum_i a_i}})$ $\implies$ $\frac{\sum_i a_i x_i}{\sum_i a_i} \geq (\Pi x_i^{a_i})^{\frac{1}{\sum_i a_i}}$
The last inequality is also the genereal weighted AM-GM inequality.
Saket Choudhary
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It's just AM-GM: $$ax+by\geq x^ay^b.$$ Done!
Michael Rozenberg
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Is it "just AM-GM" or is it rather "weighted AM-GM in two variables"? – Angina Seng Apr 22 '17 at 04:46
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@Lord Shark the Unknown AM-GM it's the following. Let $x_i>0$, $\alpha_i>0$ and $\sum\limits_{i=1}^n\alpha_i=1$. Prove that $\sum\limits_{i=1}^n\alpha_ix_i\geq\prod\limits_{i=1}^nx_i^{\alpha_i}$. In our case $n=2$. – Michael Rozenberg Apr 22 '17 at 05:42
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That's the weighted AM-GM in $n$ variables. The plain AM-GM is the case $\alpha_i=1/n$. – Angina Seng Apr 22 '17 at 05:46
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Of course the essence is correct, but a novice when looking up AM-GM might see $(x_1+\cdots +x_n)/n\ge(x_1\cdots x_n)^{1/n}$ and wonder what that has to do with the problem at hand. – Angina Seng Apr 22 '17 at 05:58
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@Lord Shark the Unknown I think it's better to write AM-GM in the form, which I wrote. I think it's very useful. – Michael Rozenberg Apr 22 '17 at 06:06
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