# I want help knowing if my solution is correct

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.

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

$$x+y=xy$$

$$AM \ge GM$$ 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?

• 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) Commented Jun 11 at 7:36
• The quadratic equation you arrived at is very misterious. Commented Jun 11 at 7:39
• @geetha290krm Ok, yes, I get this now. But I still don't understand how then the inequality still works out...? Commented Jun 11 at 7:40
• @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. Commented Jun 11 at 7:42
• 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. Commented Jun 11 at 8:07

The "Solving which gives the quadratic equation" line assumes $$x=y$$, which is not necessarily the case.
• 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. Commented Jun 12 at 3:22
• Basically, $\frac{(x+y)}{2} \ge \sqrt xy$ square on both sides and take $x$ as $4-y$ Commented Jun 12 at 3:23