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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

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    Somewhere in your calculation the RHS changed from $\sqrt{x^2+1} - x$ to $\sqrt{x^2+1} - 1$. And the last inequality is obviously false for $t = 1$. – 2017-01-26
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    Thanks Martin I have edited my post. – 2017-01-26

2 Answers 2

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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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$$\ln(t)\geq \frac{t-1}{t+1} \text{ for } t \geq 1$$

is equivalent to:

$$\ln \frac{1+u}{1-u}\ge u \text{ for }u\ge 0$$

Actually, we can prove that

$$\ln \frac{1+u}{1-u}\ge 2u$$

by noting that we can rewrite each side as

$$\int_0^u \frac{2\,dt}{1-t^2}\ge \int_0^u 2\, dt$$

Note that $2u\ge u$, and you're done!