# Proof an inequality

I'm trying to prove that $$\frac{3-2\sqrt{1-15 m^2}}{1+12 m^2}\geq 1+3 m^2$$

I have obtained in a CAS software the Taylor expansion in $m=0$

One posibility to prove the inequality is showing coeficients in Taylor expansion are non-negative, by I don't find how.

Really I want only to obtain inequality. Some idea?

EDIT

$m$ must be between $0<m<\frac{1}{\sqrt 15}$

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What is the range of $m$? –  user31373 Jun 23 '12 at 23:41
@Leonid Kovalev: I put the range in the edit, sorry. –  H. Kabayakawa Jun 23 '12 at 23:44

$$\frac{3-2\sqrt{1-15m^2}}{1+12m^2} \geq 1+3m^2 \iff$$ $$3-2\sqrt{1-15m^2} \geq (1+3m^2)(1+12m^2) \iff$$ $$2\sqrt{1-15m^2} \leq 3-(1+3m^2)(1+12m^2) \iff$$ $$\sqrt{1-15m^2} \leq \frac{3-(1+3m^2)(1+12m^2)}{2} \iff$$ $$\sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2}$$ Note that on the interval you're concerned about, the right hand side is always positive. Proof: it's obviously decreasing on $\left(0,\frac{1}{\sqrt{15}}\right)$, and is equal to $\frac{21}{50}$ at the right endpoint. Therefore squaring both sides is legal here with an $\iff$ statement. $$\sqrt{1-15m^2} \leq \frac{2-15m^2-36m^4}{2} \iff$$ $$1-15m^2 \leq \left(\frac{2-15m^2-36m^4}{2}\right)^2 \iff$$ $$1-15m^2 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4 - 15m^2 + 1 \iff$$ $$0 \leq 324m^8 + 270m^6 + \frac{81}{4}m^4$$

This last statement is clearly true.

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Multiply by the denominator $1+12m^2$. You get $1+15m^2+36m^4$ on the right. On the left side, $3-2\sqrt{1-15m^2}=1+15m^2+(225/4)m^4+\dots$ where all higher order terms have positive coefficients by the binomial series formula (the expansion of $(1+x)^p$) .

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I suspect you derived your expression from the lowest root of the quadratic polynomial $$P(x)=(1+12m^2)x^2-6x+5$$ Because the leading coefficient is positive and $1+3m^2$ is less than the mean of the roots $3/(1+12m^2)$ when $3m^2\le 1/5$, this root is no less than $1+3m^2$ if and only if $$P(1+3m^2)\ge 0$$ Letting $3m^2=t$: $$P(1+t)=(1+4t)(1+t)^2-6(1+t)+5=4t^3+9t^2\ge 0$$

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