I am working on a howework question, trying to prove the following:

$$5a+b > 4\sqrt{ab},$$

where $a$ and $b$ are positive real numbers.

I've tried multiplying expression by $\sqrt{ab}$, squaring both sides of the equation so far. In both cases, after re-factoring, I could not conclude that inequality holds. Can someone point me in the right direction?

  • 3
    $\begingroup$ Use $\frac{5a+b}{2} \geq \sqrt{5ab}$ and also the fact that $\sqrt{5} > 2$ to get the desired result. $\endgroup$ – Kirthi Raman May 13 '12 at 22:54

In general you have to be a bit careful about squaring inequalities, but here there’s no problem: since we’re assuming that $a$ and $b$ are positive, $5a+b$ is positive, and therefore $5a+b>4\sqrt{ab}$ if and only if $(5a+b)^2>16ab$, i.e., if and only if $25a^2-6ab+b^2>0$. Now think of $b^2-6ab+25a^2$ as a quadratic in $b$ and complete the square: $b^2-6ab+25a^2=(b-3a)^2+16a^2$. Clearly $(b-3a)^2\ge 0$, and since $a$ is positive, $a^2>0$, so $(b-3)^2+16a^2>0$.

Retracing our steps, we see that $(5a+b)^2>16ab$ and hence that $5a+b=\sqrt{(5a+b)^2}>4\sqrt{ab}$.

  • $\begingroup$ You mean $b-3a$, not $b-3$, I think. $\endgroup$ – Cameron Buie May 13 '12 at 22:59
  • $\begingroup$ @Cameron: I sure do; thanks! $\endgroup$ – Brian M. Scott May 13 '12 at 23:00

For any real $x$ and $y$ you have $$2xy=x^2+y^2-(x-y)^2$$ taking $x=2\sqrt{a}$ and $y=\sqrt{b}$ you get $$2(2\sqrt{a}\sqrt{b})=4 a+b-(2\sqrt{a}-\sqrt{b})^2\leq 4a+b < 5a+b$$ the only remaining case (i.e. $a=0$) is trivial.


Let $a=x^2$ and $b=y^2$, where $x$ and $y$ are positive. We want to show that $5x^2-4xy+y^2 \gt 0$. This is clear from the fact that already $4x^2-4xy+y^2=(2x-y)^2 \ge 0$.

Alternately, note that $5x^2-4xy+y^2=x^2((y/x)^2-4(y/x)+5)$. Let $t=y/x$, and use the fact that $t^2-4t+5$ is positive at (say) $t=0$, and is never $0$, so it is always positive.


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