# Dan Sitaru's Inequality with a Double Integral

### Solution

If $a,b,m,n\gt 0$ then

$(m^2\sqrt{mnab}+a^2)(n^2\sqrt{mnab}+b^2)\ge (mb+na)^2\sqrt{mnab}.$

For a proof, let $f:\,\mathbb{R}\to (0,\infty),$ defined by $f(x)=(m^2x^2+a^2)(n^2x^2+b^2).$ Then

\begin{align} f(x)&=(m^2x^2+a^2)(n^2x^2+b^2)\\ &=(mx-ai)(mx+ai)(nx-bi)(nx+bi)\\ &=[(mnx^2-ab)+i(mb+na)x][(mnx^2-ab)-i(mb+na)x]\\ &=(mnx^2-ab)^2+(mb+na)^2x^2\\ &\ge (mb+na)^2x^2. \end{align}

It follows that

$(m^2x^2+a^2)(n^2x^2+b^2)\ge (mb+na)^2x^2.$

Substitute $x=\sqrt[4]{mnab}:$

$(m^2\sqrt{mnab}+a^2)(n^2\sqrt{mnab}+b^2)\ge (mb+na)^2\sqrt{mnab}$

and note that the equality holds when $m^3n^3=ab.$

To continue with the original problem, set $a=f(x)$ and $b=f(y):$

\begin{align} &\left(m^2\sqrt{mnf(x)f(y)}+f^2(x)\right)\left(n^2\sqrt{mnf(x)f(y)}+f^2(y)\right)\\ &\qquad\qquad\qquad\ge (mf(y)+nf(x))^2\sqrt{mnf(x)f(y)}\\ &\qquad\qquad\qquad\ge (mf(y)+nf(x))^2\sqrt{mn}\end{align}

so that

$\small{\sqrt{\left(m^2\sqrt{mnf(x)f(y)}+f^2(x)\right)\left(n^2\sqrt{mnf(x)f(y)}+f^2(y)\right)}\ge (mf(y)+nf(x)})\sqrt[4]{mn}.$

Now integrate:

\displaystyle\begin{align} &\small{\int_0^1\int_0^1\sqrt{\left(m^2\sqrt{mnf(x)f(y)}+f^2(x)\right)\left(n^2\sqrt{mnf(x)f(y)}+f^2(y)\right)}dxdy}\\ &\qquad\qquad\qquad\ge \small\sqrt[4]{mn}{\int_0^1\int_0^1(mf(y)+nf(x))dxdy=\sqrt[4]{mn}(m+n)\int_0^1f(x)dx}. \end{align}

### Acknowledgment

This problem was kindly communicated to me by Dan Sitaru, along with a solution of his.