calc inequality

**BenWong** · November 20th 2007, 11:57 AM

My books says to use calculus and algebra to prove this inequality, assuming that x is a positive number:
ln(1+x) less than or equal to x/(sqrt(1+x)).
I have no idea what to do and would greatly appreciate any help. Thanks in advance.

**PaulRS** · November 20th 2007, 12:20 PM

Consider: $\frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x+1)\cdot{\sq rt[]{x+1}}}}}$

Which clearly holds for $x\geq{0}$

Just to see: $\frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x+1)\cdot{\sq rt[]{x+1}}}}}$

If and only if (since $x\geq{0}$ ) : $1\leq{\frac{x+2}{2\cdot{\sqrt[]{x+1}}}}$
Which is:
$1\leq{\frac{\sqrt[]{x+1}}{2}+\frac{1}{2\cdot{\sqrt[]{x+1}}}}$

Let: $z=\sqrt[]{x+1}\geq{1}$

So: $1\leq{\frac{z}{2}+\frac{1}{2\cdot{z}}}$

Multiplying by z: $z\leq{\frac{z^2}{2}+\frac{1}{2}}$ that is true since (z-1)^2>=0 and then z²+1>=2z

Integrating the inequality: $\int_0^b\frac{dx}{x+1}\leq{\int_0^b\frac{x+2}{2\cd ot{(x+1)\cdot{\sqrt[]{x+1}}}}}dx$ for $b\geq{0}$ What do you see?

**BenWong** · November 20th 2007, 12:30 PM

Is it really this difficult? We haven't gone over integrals, so I'm not really sure I understand the idea, but thanks for trying to help me. Is there something simpler (possibly involving differentation)?

**PaulRS** · November 20th 2007, 12:35 PM

Both expressions are equal when x=0, but their derivatives have the following property: $(\ln(x+1))'=\frac{1}{x+1}\leq{\frac{x+2}{2\cdot{(x +1)\cdot{\sqrt[]{x+1}}}}}=(\frac{x}{\sqrt[]{1+x}})'$
(as shown in the other post)
for $x\geq{0}$

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