How would you prove the following inequality using the Mean Value Theorem:
$$1+2\ln x\leq x^2$$ for $x>0$.
Answer
Let $f (t)=1+2\ln(t)-t^2$ for $t>0$.
for $x>0$ , $f $ is continuous at $[x,1] $ or $[1,x]$ and differentiable at $(x,1) $ or $(1,x) $ thus by MVT, exists $c$ strictly between $x $ and $1$ such that
$$f (x)-f (1)=(x-1)f'(c) $$
$$1+2\ln (x)-x^2=2(x-1)(1/c -c) $$
$$=2 (x-1)\frac {1-c^2}{c} $$
If $x>1$ then $1-c^2<0$ and
if $x <1$ then $1-c^2>0$
thus in all cases
$$f (x)\le 0.$$
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