Prove $\frac{1}{2} + \frac{1}{2(u+1)^2} - \frac{1}{\sqrt{1+2u}} \geq 0$ for $u \geq 0$

Question

This inequality provides a tight lower bound to $\sqrt{1+2u}$ for $u\geq 0$ 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.

Answer 1

It's $$\sqrt{2u+1}(u^2+2u+2)\geq2(u+1)^2$$ or $$(2u+1)(u^2+2u+2)^2\geq4(u+1)^4$$ or $$u^3(2u^2+5u+4)\geq0.$$ Done!

Also, we can use the following way.

Let $\sqrt{2u+1}=x$.

Thus, $x\geq1$ and we need to prove that $$\frac{1}{2}+\frac{1}{2\left(\frac{x^2-1}{2}+1\right)^2}\geq\frac{1}{x}$$ or $$1+\frac{4}{(x^2+1)^2}\geq\frac{2}{x}$$ or $$x(x^2+1)^2+4x\geq2(x^2+1)^2$$ or $$x^5+2x^3+x+4x\geq2x^4+4x^2+2$$ or $$x^5-2x^4+2x^3-4x^2+5x-2\geq0$$ or $$x^5-2x^4+x^3+x^3-2x^2+x-2x^2+4x-2\geq0$$ or $$(x-1)^2(x^3+x-2)\geq0,$$ which is obvious for $x\geq1$.

Done againe!

Answer 2

write your inequality in the form $$\frac{1}{2}+\frac{1}{2(u+1)^2}\geq \frac{1}{\sqrt{1+2u}}$$ and square it after this we get $$1/4\,{\frac {{u}^{3} \left( 5\,u+2\,{u}^{2}+4 \right) }{ \left( u+1 \right) ^{4} \left( 1+2\,u \right) }} \geq 0$$