<s>[INST] Prove that $1+x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}\;\forall x \geq 0$ without derivative 
Prove that $1+x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}\;\forall x \geq 0$

$\bf{My\; Try::}$ Means we have to prove $x\ln(x+\sqrt{x^2+1})\geq \sqrt{1+x^2}-1$
Put $\sqrt{x^2+1}+x = t\geq 1\;,$ Then $\displaystyle (\sqrt{x^2+1}-x) = \frac{1}{t}.$ 
So we get  $\displaystyle x = \frac{t^2-1}{2t}$ and $\displaystyle \sqrt{x^2+1} = \frac{t^2+1}{2t}$
So we have to prove $\displaystyle \left(\frac{t^2-1}{2t}\right)\ln (t)\geq \frac{t^2+1-2t}{2t}\Rightarrow (t^2-1)\ln t \geq (t-1)^2$
So we have to prove $\displaystyle \ln(t)\geq \frac{t-1}{t+1}\forall t \geq 1$
Help required for proving above statement, Thanks
 [/INST] Applying the "well-known" inequality $\ln y \le y - 1$ 
to $y = \frac 1t $ gives for $t \ge 1$
$$
 \ln t = - \log \frac 1t \ge -(\frac 1t-1) = \frac{t-1}{t} \ge  \frac{t-1}{t+1} \, .
$$
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