How do I show this inequality:

For positive real numbers $a,b$ such that $a+b=1$ then $\frac{2}{\frac{a}{x}+ \frac{b}{y}} \leq ax + by$ for $x, y > 0$


closed as off-topic by Macavity, graydad, user147263, Najib Idrissi, apnorton Feb 28 '15 at 19:40

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  • $\begingroup$ Are you sure the LHS has numerator $2$ and not $1$? Your inequality does not hold for say $a=b=x=y=\frac12$. $\endgroup$ – Macavity Feb 28 '15 at 16:17

The $2$ in your inequality should be $1$. Indeed,

$$(ax + by)\left(\frac{a}{x} + \frac{b}{y}\right) = a^2 + b^2 + ab\left(\frac{x}{y} + \frac{y}{x}\right) \ge a^2 + b^2 + 2ab\sqrt{\frac{x}{y}\frac{y}{x}} = a^2 + b^2 + 2ab = (a + b)^2 = 1,$$

with equality if and only if $x = y$.

  • $\begingroup$ You can apply the Cauchy-Schawarz Inequality right away, but anyway nice proof. $\endgroup$ – Stefan4024 Feb 28 '15 at 17:00
  • 1
    $\begingroup$ @Stefan4024 yes I know that, but I figured this way might be easier to follow. $\endgroup$ – kobe Feb 28 '15 at 17:03

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