A certain inequality involving square roots of polynomials

Question

I want to prove the inequality $$\begin{aligned} &\sqrt{(x - 1)^2 + y^2}\Big[y^2(9x - 6) - 9x^2 + 9x^3\Big]+ y^2(16x^2 - 16x + 7)\\ &- \sqrt{x^2 + y^2}\Big[9x + y^2(9x - 3) + \sqrt{(x - 1)^2 + y^2}(9x^2 - 9x + 6y^2) - 18x^2 + 9x^3\Big]\\ & + 9x^2 - 18x^3 + 9x^4 + 7y^4 \geq 0 \end{aligned} $$ for real and nonzero $x,y$. Is this inequality true? How does one go about showing it?

Answer 1

A human verifiable proof.

The desired inequality is written as $$p A + B - q(C + p D) \ge 0 \tag{1}$$ where \begin{align*} &p := \sqrt{(x-1)^2 + y^2}, \quad q := \sqrt{x^2 + y^2}, \quad A := y^2(9x - 6) - 9x^2 + 9x^3, \\ &B := y^2(16x^2 - 16x + 7) + 9x^2 - 18x^3 + 9x^4 + 7y^4, \\ &C := 9x + y^2(9x - 3) - 18x^2 + 9x^3, \quad D := 9x^2 - 9x + 6y^2. \end{align*}

Clearly $B \ge 0$. We claim that $pA + B \ge 0$. Indeed, we have \begin{align*} B^2 - p^2 A^2 &= 49\,{y}^{8}+ \left( 143\,{x}^{2}-116\,x+62 \right) {y}^{6} \\ &\qquad + \left( 139 \,{x}^{4}-224\,{x}^{3}+165\,{x}^{2}-44\,x+13 \right) {y}^{4} \\ &\qquad + 9x^2(5x^2-2x+2)(x-1)^2{ y}^{2}\\ &\ge 0. \end{align*} (Note: All the coefficients, as a polynomial in $y$, are non-negative.) The claim is proved.

To prove (1), it suffices to prove that $(pA + B)^2 - q^2(C + pD)^2 \ge 0$, or $$Mp + N \ge 0 \tag{2}$$ where \begin{align*} M &:= 18\,{x}^{5}{y}^{2}+36\,{x}^{3}{y}^{4}+18\,x{y}^{6}-90\,{x}^{4}{y}^{2}- 138\,{x}^{2}{y}^{4}-48\,{y}^{6}\\ &\qquad +144\,{x}^{3}{y}^{2}+156\,x{y}^{4}-72\, {x}^{2}{y}^{2}-84\,{y}^{4},\\ N &:= 13\,{y}^{8}+ \left( 44\,{x}^{2}-98\,x+89 \right) {y}^{6}\\ &\qquad + \left( 49\,{ x}^{4}-206\,{x}^{3}+327\,{x}^{2}-242\,x+85 \right) {y}^{4} \\ &\qquad + 18x^2(x-1)^2(x-2)^2{ y}^{2}. \end{align*}

Note that if $\sqrt{(x-1)^2 + y^2} = 0$, then $x=1, y= 0$ and $Mp + N = 0$. In what follows, assume that $\sqrt{(x-1)^2+y^2} > 0$. We use the parametrization $$x := 1 + r \cdot \frac{2t}{1+t^2}, \quad y := r \cdot \frac{1-t^2}{1+t^2}, \quad r > 0, t \in \mathbb{R}.$$ We have $p = \sqrt{(x-1)^2+y^2} = r$ and \begin{align*} Mp + N &= \frac{r^6(t - 1)^2(t + 1)^4(13t^2 + 10t + 13)}{(t^2+1)^4}\cdot g(r, t) \end{align*} where \begin{align*} g(r, t) &:= \left(r + \frac{1}{r} - \frac{5(3t^2+2t+3)(t-1)^2}{(t^2+1)(13t^2+10t+13)}\right)^2\\[6pt] &\qquad - \frac{36(t+3)(3t+1)(t+1)^2(t^2+t+1)^2}{(13t^2+10t+13)^2(t^2+1)^2}. \end{align*} It suffices to prove that $g(r, t)\ge 0$ for all $r > 0$ and $t \in \mathbb{R}$ (easy, by letting $u := r + 1/r \ge 2$; we only need to consider the case $(t+3)(3t+1) > 0$).

We are done.

Answer 2

This problem is one of real algebraic geometry. As such, it admits a completely algorithmic solution. In Mathematica, such algorithms are realized by commands such as Reduce. Using this command, one finds in about 0.05 sec that (i) the left-hand side of your inequality is $\ge0$ for all real nonzero $x,y$ and (ii) the left-hand side of your inequality is $0$ for real nonzero $x,y$ iff $x=1/2$ and $y=\pm\sqrt3/2$.

Here is an image of the corresponding Mathematica notebook:

Answer 3

My second solution (simpler than my first solution).

Remark. Inspired by @Toni Mhax's solution, we came up with this solution.

Let $a := \sqrt{(x-1)^2 + y^2},\, b := \sqrt{x^2 + y^2}$. We have $x = \frac{b^2 - a^2 + 1}{2}, \, y^2 = b^2 - \big(\frac{b^2 - a^2 + 1}{2}\big)^2$. The desired inequality is written as $$ \frac{y^2}{a + b + 1} F(a, b)\ge 0, \tag{1}$$ where $F(a, b) := -b^3 + (2a+2)b^2 + (2a^2-9a+2)b-a^3+2a^2+2a-1$. Also, from $a\ge 0$ and $b^2 - \big(\frac{b^2 - a^2 + 1}{2}\big)^2 \ge 0$, we have $|a-1| \le b \le a + 1$.

We can prove that $$|a-1| \le b \le a + 1 \implies F(a, b) \ge 0. \tag{2}$$ A proof of (2) is given at the end which is easy but not so nice. Hope to see a nice proof of (2).

Thus, we are done.

$\phantom{2}$

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We split into two cases.

• Case 1. $\frac{13}{25} \le a \le \frac{19}{10}$

If $a = 1$, we have $F(a, b) = (2-b)(b-1)^2 \ge 0$.

If $a \ne 1$, as a cubic in $b$, the discriminant of $F$ is given by $$\Delta_b(F) = (125a^4-550a^3+847a^2-550a+125)(a-1)^2 < 0.$$ Also, we have $F(a, a+1) = 2(a+1))(a-1)^2 > 0$. Thus, $F(a, b) \ge 0$ for all $b \le a + 1$.

• Case 2. $(0 \le a < \frac{13}{25}) \lor (a > \frac{19}{10})$

We have $\frac{\partial F}{\partial b} = -3b^2 + (4a+4)b + 2a^2-9a+2 \ge 0$ for all $|a-1| \le b \le a + 1$ (easy). Thus, we have $$F(a, b) \ge F(a, |a-1|) = \left\{\begin{array}{ll} 2(2a-1)^2 & 0 \le a \le 1 \\ 2a(a-2)^2 & a > 1. \end{array} \right. \ge 0.$$

In summary, (2) is true.

Answer 4

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third-degree polynomial $p$, the derivative has one negative and one positive root, so we need only to check $p(x)\ge 0$ when $0\le x\le a$, (for example, $-5a^4-20a^3+47a^2-20a-5 \le -5a^4-20a^3+50a^2-20a-5$ and this is $-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0; a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $$t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0.$$

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).