If $a, b, x, y$ are positive rational numbers such that $\frac 1x + \frac 1y = 1$ then prove that $\frac {a^{x}}{x}+ \frac {b^{y}}{y}$ $\ge ab$

This question is from Problems Plus in IIT Mathematics. I tried to solve it with a different method than the one given, and I need to know if it is correct.

$\frac 1x + \frac 1y = 1$


$AM \ge GM$ for all positive numbers, so $\frac{(x+y)}{2} \ge \sqrt{xy}$

$=\frac{xy}{2} \ge \sqrt{xy}$ since $x+y=xy$

Solving which gives the quadratic equation $-y^{2}+4y-4=0$

$y=2$ and $x=4-2=2$

To prove: $\frac {a^{x}}{x}+ \frac {b^{y}}{y}$ $\ge ab$

$AM \ge GM$

$\frac{a^{2}}{2}+\frac{b^{2}}{2} \ge ab$

$a^{2} + b^{2} \ge 2ab$

$a^{2}+b^{2}-2ab \ge 0$

$(a-b)^{2} \ge 0$

We know that this holds true for all positive values of $a$ and $b$, hence proved.

Is this correct? If not, where did I go wrong?

  • 2
    $\begingroup$ You canot find the exact values of $x$ and $y$ from the equation $\frac 1 x+\frac 1 y=1$. ($x=4, y=4/3$ is another solution) $\endgroup$ Commented Jun 11 at 7:36
  • 2
    $\begingroup$ The quadratic equation you arrived at is very misterious. $\endgroup$ Commented Jun 11 at 7:39
  • $\begingroup$ @geetha290krm Ok, yes, I get this now. But I still don't understand how then the inequality still works out...? $\endgroup$
    – Phoenix
    Commented Jun 11 at 7:40
  • $\begingroup$ @geetha290krm Right I got where I went wrong. The question doesn't say x and y are in an AP so I can't use the inequality in the first place. I'm so sorry lol. $\endgroup$
    – Phoenix
    Commented Jun 11 at 7:42
  • $\begingroup$ The easiest way to see this is that $\frac{a^x}{x}+\frac{b^y}{y}$ is a convex combination sinc $\frac{1}{x}+\frac{1}{y}=1$ and the natural logarithm is a concave function. $\endgroup$ Commented Jun 11 at 8:07

1 Answer 1


The "Solving which gives the quadratic equation" line assumes $x=y$, which is not necessarily the case.

  • $\begingroup$ Not really. I why my solution is incorrect, which is basically step 1 using AM-GM inequality for two numbers that are not necessarily in an AP. I didn't really assume $x=y$, but I skipped some steps I thought were obvious. $\endgroup$
    – Phoenix
    Commented Jun 12 at 3:22
  • $\begingroup$ Basically, $\frac{(x+y)}{2} \ge \sqrt xy$ square on both sides and take $x$ as $4-y$ $\endgroup$
    – Phoenix
    Commented Jun 12 at 3:23

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