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Solve in $\mathbb{R}$:

$$ \frac{3x+3}{\sqrt{x}}=4+\frac{x+1}{\sqrt{x^{2}-x+1}} $$

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What did you try? –  Listing Jun 22 '12 at 14:59
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$${}{}{}{}1{}{}{}{}$$ –  The Chaz 2.0 Jun 22 '12 at 15:03
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1 Answer

up vote 9 down vote accepted

Note that $\frac{3x+3}{\sqrt{x}}\ge 6$, with equality only at $x=1$. This comes down to the inequality $(\sqrt{x}-1)^2 \ge 0$.

Or else we can simply quote the fact that if $t$ is positive then $t+\frac{1}{t}\ge 2$.

Now look at the right-hand side. It is not hard to show, using calculus or, more easily, without using calculus, that for positive $x$ $$4+\frac{x+1}{\sqrt{x^2-x+1}}\le 6,$$ with equality only at $x=1$. This comes down to showing that $\frac{x+1}{\sqrt{x^2-x+1}}\le 2$, or equivalently $(x+1)^2 \le 4(x^2-x+1)$, that is, $3(x^2-2x+1) \ge 0$.

So the only real solution is $x=1$.

Remark: Or else we could start squaring and rearranging and perhaps squaring again. Not an appealing prospect! It would likely turn something that has a reasonably nice shape into a mess.

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