<s>[INST] How to prove that $\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$? $$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$$
Equality seems to be when x = 1
I have managed to show that the derivative is 0 at x = 1, and that this is a minimum (by the second derivative test), but I am stuck on how to show that this is the only minimum.
If $f(x)$ is defined as the LHS, then we have
$$\begin{array}{l}f\left(x\right)=\sqrt{\frac{x^2-1+2}{x+1}}+\frac{2}{\sqrt{x}+1}=\sqrt{x-1+\frac{2}{x+1}}+\frac{2}{\sqrt{x}+1}\\
f^{\prime}\left(x\right)=\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}-\frac{2}{\left(\sqrt{x}+1\right)^2}\cdot\frac{1}{2\sqrt{x}}=0\\
\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\\
\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\left(x+1\right)^2-2\right)=2\left(x+1\right)^2\sqrt{\frac{x^2+1}{x+1}}\\
x\left(x+1+2\sqrt{x}\right)^2\left(\left(x+1\right)^2-2\right)^2=2\left(x+1\right)^4\left(\frac{x^2+1}{x+1}\right)\end{array}$$
At this point the computations become unrealistic without enlisting the help of wolfram alpha to expand everything. I did this, and was left with a massive polynomial which had the root x = 1.
Here is a graph
 [/INST] Apply Cauchy-Schwarz inequality twice followed by an application of AM-GM inequality:
$\sqrt{\dfrac{x^2+1}{x+1}}+\dfrac{2}{\sqrt{x}+1}\ge \dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{x}+1}\ge\dfrac{\sqrt{x+1}}{\sqrt{2}}+\dfrac{2}{\sqrt{2(x+1)}}\ge 2\sqrt{\dfrac{\sqrt{x+1}}{\sqrt{2}}\cdot \dfrac{2}{\sqrt{2(x+1)}}}=2$. Thus the minimum value is $2$ and this is achieved when $x = 1$.
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