# Prove 1/2 + (1)/(2(u+1)^2) - 1/(sqrt(1+2u)) >= 0 for u>=0

Prove $\frac{1}{2}+\frac{1}{2\left(u+1{\right)}^{2}}-\frac{1}{\sqrt{1+2u}}\ge 0$ for $u\ge 0$
This inequality provides a tight lower bound to $\sqrt{1+2u}$ for $u\ge 0$ without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder if there is a quicker solution. Thanks.
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wlanauee
It's
$\sqrt{2u+1}\left({u}^{2}+2u+2\right)\ge 2\left(u+1{\right)}^{2}$
or
$\left(2u+1\right)\left({u}^{2}+2u+2{\right)}^{2}\ge 4\left(u+1{\right)}^{4}$
or
${u}^{3}\left(2{u}^{2}+5u+4\right)\ge 0.$
Done!
Also, we can use the following way.
Let $\sqrt{2u+1}=x$
Thus, $x\ge 1$ and we need to prove that
$\frac{1}{2}+\frac{1}{2{\left(\frac{{x}^{2}-1}{2}+1\right)}^{2}}\ge \frac{1}{x}$
or
$1+\frac{4}{\left({x}^{2}+1{\right)}^{2}}\ge \frac{2}{x}$
or
$x\left({x}^{2}+1{\right)}^{2}+4x\ge 2\left({x}^{2}+1{\right)}^{2}$
or
${x}^{5}+2{x}^{3}+x+4x\ge 2{x}^{4}+4{x}^{2}+2$
or
${x}^{5}-2{x}^{4}+2{x}^{3}-4{x}^{2}+5x-2\ge 0$
or
${x}^{5}-2{x}^{4}+{x}^{3}+{x}^{3}-2{x}^{2}+x-2{x}^{2}+4x-2\ge 0$
or
$\left(x-1{\right)}^{2}\left({x}^{3}+x-2\right)\ge 0,$
which is obvious for $x\ge 1$.
Done againe!
###### Did you like this example?
Gisselle Hodges
write your inequality in the form
$\frac{1}{2}+\frac{1}{2\left(u+1{\right)}^{2}}\ge \frac{1}{\sqrt{1+2u}}$
and square it after this we get
$1/4\phantom{\rule{thinmathspace}{0ex}}\frac{{u}^{3}\left(5\phantom{\rule{thinmathspace}{0ex}}u+2\phantom{\rule{thinmathspace}{0ex}}{u}^{2}+4\right)}{{\left(u+1\right)}^{4}\left(1+2\phantom{\rule{thinmathspace}{0ex}}u\right)}\ge 0$