Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $

Question

As the title described, I was trying to find an alternative proof to

If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.

Here's the proof that I've found (I'm sorry, I forgot where I got it):

Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.

Now since I love to punish myself, I tried to find a harder proof as such:

We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l } \cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\ (\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\ (\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\ \cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\ \dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\ \end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1 $$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$

Now how do I prove the sextic polynomial inequality above (which is true)?

Answer 1

(This has a fairly obvious improvement of the inequality.)

Here is a simple algebraic proof of the stronger inequalities $x^2+y^2=1 \implies x^3+y^3 \ge \dfrac{x+y}{2} \ge \sqrt{xy}.$

More generally, without the restriction on $x^2+y^2$, this gives $x^3+y^3 \ge \dfrac{(x+y)(x^2+y^2)}{2} $.

Proof that this inequality is stronger:

$xy \le (x^2+y^2)/2 = 1/2$ so $1/\sqrt{xy} \ge \sqrt{2}$. Therefore $\sqrt{xy} = xy/\sqrt{xy} \ge xy \sqrt{2}$.

Proof of the inequality.

$\begin{array}\\ x^3+y^3 &=(x+y)(x^2-xy+y^2)\\ &= (x+y) \dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\ &= (x+y) \dfrac{1+(x-y)^2}{2}\\ &\ge \dfrac{x+y}{2}\\ &\ge \sqrt{xy}\\ \end{array} $

In all these inequalities, there is equality when x=y and strict inequality otherwise.

Example.

If $x=3/5, y=4/5$ then

$x^3+y^3 = (27+64)/125 =91/125=0.728,\\ \dfrac{x+y}{2} = 7/10 =0.7\\ \sqrt{xy} = \sqrt{12/25} = 2 \sqrt{3}/5 = 0.6928...,\\ xy \sqrt{2} = 0.6788... $

Answer 2

An alternative way of proving, is going into parametric equations. The cubic equation is a Follium of Descartes, and can be represented by $(\frac{\sqrt{2}t}{1+t^3},\frac{\sqrt{2}t^2}{1+t^3})$. With respect to the 1st quadrant we must now show that all points on the unit circle fall "outside" (that is on the "right" side) of the Follium. Note that the point$P(\sqrt{2}/2,\sqrt{2}/2)$ coincides with both curves, hence the need for the equality symbol. When we use the distance formula on the parametric representation (in squared form), we get $d^2=\frac{2t^2+2t^4}{(1+t^3)^2}$. We can optimize $d^2$ and show that for $t=1$, a local maximum is achieved for the distance. You need to apply the quotient rule, which is annoying algebra. The numerator will be $(1+t^3)(-4t^6-8t^4+8t^3+4t)$ and when you set this equal to zero, $t=-1$ doesn't yield anything productive (what's going on with $t=-1$?). But the sixth degree polynomial inside has a root $t=0$ (Origin) and $t=1$ (Rational Zero Theorem!) and thus is divisible by $(t-1)$, which (after factoring out $-4t$) gives $t^4+t^3+3t^2+t+1$ which has no real roots. It shouldn't be to hard to show that for $t=1$ a maximum distance (Point P) in the first quadrant is achieved. Here you have all the ingredients to set up the proof which is what I want you to do.

Answer 3

After you substitute $(x,y)=(\cos{\theta},\sin{\theta})$, you just need to prove the followings:

$$ \begin{aligned} \sin{\theta}+\cos{\theta}&\geq\sqrt{2}\\ \\ 1-\sin{\theta}\cos{\theta}&\geq\sin{\theta}\cos{\theta} \end{aligned} $$

Which are easy. First one is well known, second one is just double angle sine identity

Answer 4

HINT

Since $x > 0$ and $y > 0$, the proposed inequality is equivalent to

\begin{align*} x^{3} + y^{3} \geq \sqrt{2}xy & \Longleftrightarrow (x^{3} + y^{3})^{2} \geq 2x^{2}y^{2}\\\\ & \Longleftrightarrow (x+y)^{2}(1 - xy)^{2} \geq 2x^{2}y^{2}\\\\ & \Longleftrightarrow (1 + 2xy)(1-xy)^{2} \geq 2x^{2}y^{2}\\\\ \end{align*}

If we make the change of variable $t = xy$, we obtain the following equivalent inequality: \begin{align*} (1 + 2t)(1 - t)^{2} \geq 2t^{2} & \Longleftrightarrow (1+2t)(1 - 2t + t^{2})\geq 2t^{2}\\\\ & \Longleftrightarrow 1 - 2t + t^{2} + 2t - 4t^{2} + 2t^{3} \geq 2t^{2}\\\\ & \Longleftrightarrow 2t^{3} - 5t^{2} + 1 \geq 0\\\\\ & \Longleftrightarrow (2t^{3} - t^{2}) - (4t^{2} - 1) \geq 0\\\\ & \Longleftrightarrow t^{2}(2t - 1) - (2t-1)(2t+1) \geq 0\\\\ & \Longleftrightarrow (2t-1)(t^{2} - 2t - 1) \geq 0 \end{align*}

Can you take it from here?

Answer 5

$(x+y)^2 =x^2+2xy+y^2 =2xy+1 $ so $x+y =\sqrt{2xy+1} $.

$\begin{array}\\ x^3+y^3 &=(x+y)(x^2-xy+y^2)\\ &=\sqrt{2xy+1}\dfrac{2x^2-2xy+2y^2}{2}\\ &=\sqrt{2xy+1}\dfrac{x^2+y^2+x^2-2xy+y^2}{2}\\ &=\sqrt{2xy+1}\dfrac{1+(x-y)^2}{2}\\ &\ge\dfrac12\sqrt{2xy+1}\\ \end{array} $

so we want

$\dfrac12\sqrt{2xy+1} \ge xy\sqrt{2} $ or $2xy+1 \ge 8(xy)^2 $ or, with $z = xy$, $8z^2-2z-1 \le 0 $.

$8z^2-2z-1 =8(z+\frac14)(z-\frac12) $.

$0 \le (x-y)^2 =x^2-2xy+y^2 $ so $2xy \le x^2+y^2 =1 $ or $0 \le z \le \frac12 $.

Therefore $8z^2-2z-1 =8(z+\frac14)(z-\frac12) \le 0 $

which is what we want.

Answer 6

(I agree that most of the algebraic steps here are not obvious nor intuitive. It was a lot of wishful thinking that this method could work, in part because of the background of the problem.)

For the sextic inequality, from the graph, you know that $\sqrt{2} - 1 $ is a double root, so $ ( x - (\sqrt{2} - 1) )^2$ is a factor of the polynomial.
(Alternatively, we know from the work you did that that's the equality case.)

So, we have

$$\frac{ x^6 - 2\sqrt{2} x^5 - 3 x^4 - 8 x^3 + 3 x^2 + 2 \sqrt{2} x - 1} { ( x - (\sqrt{2} - 1) )^2} \\ = x^4 - 2x^3 - 2 ( 1 + \sqrt{2} ) x^2 - 2 ( 3 + 2 \sqrt{2} ) x - 2\sqrt{2} - 3.$$

We make the observation (by staring really hard) that the quartic factors as

$$ (x^2 - ( 2 + \sqrt{2} ) x - \sqrt{2} - 1 ) ( x^2 + \sqrt{2} x + \sqrt{2} + 1 ) $$

We can now easily show that on $ 0 < x < 1$, this value is negative. (First term is negative, second term is positive.)
Hence, the original expression is $ \leq 0$, with equality at $ \sqrt{2} - 1 $.

Answer 7

Here a relatively "natural" way to show the inequality. It avoids a time consuming (or rather time wasting) analysis of a quite specific 6th degree polynomial where the amount of learnings and insights to be expected doesn't seem to justify the effort.

To show is equivalently

$$\frac{x^2}y + \frac{y^2}x \geq \sqrt 2$$

\begin{eqnarray*}\frac{x^2}y + \frac{y^2}x & \stackrel{\frac 1t \text{convex for }t>0}{\geq} & \frac 1{x^2y+y^2x}\\ & = & \frac 1{xy(x+y)}\\ & \stackrel{GM-AM: xy\leq \frac 12}{\geq} & \frac 2{x+y}\\ & \stackrel{C.S.:x+y\leq \sqrt 2}{\geq} & \frac 2{\sqrt 2} = \sqrt 2 \end{eqnarray*}

Answer 8

Alternative proof via C-S and AM-GM.

$$x^3+y^3\ge \sqrt 2 xy \iff \frac{x^2}{y} + \frac{y^2}{x} \ge \sqrt 2$$

$$1 = (x^2 + y^2)^2 \le \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \cdot (x^2 y + xy^2) \tag {CS}$$

$$= \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{x^2y^2(x+y)^2}\\ = \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{(xy)^2(1+2xy)}\\ \le \left( \frac{x^2}{y} + \frac{y^2}{x} \right) \sqrt{\frac 14 \left(1+2\frac 12\right)}\\$$

Done.

Answer 9

Another proof inspired by marty cohen.

By Power Mean Inequality,

$$\left(\frac{x^3+y^3}{2}\right)^{1/3} \ge \left( \frac{x^2+y^2}{2}\right)^{1/2}=2^{-\frac 12}\\ \implies x^3+y^3\ge 2\cdot 2^{-\frac 32}=\frac{1}{\sqrt 2} = \frac{x^2+y^2}{\sqrt 2}\ge \sqrt 2 xy.\blacksquare$$

Answer 10

By CS-Engel you will get

$x^3+y^3 \ge \dfrac{(x^2+y^2)^2}{x+y} = \dfrac{(x^2+y^2)(x^2+y^2)}{x+y} = \dfrac{x^2+y^2}{x+y}$ ... (1)

By QM-AM inequality $\dfrac{1}{\sqrt{2}} = \sqrt{\dfrac{x^2+y^2}{2}} \ge \dfrac{x+y}{2}$ or $\dfrac{1}{x+y} \ge \dfrac{1}{\sqrt{2}}$ ...(2)

Combining (1) and (2) with AM-GM we get $\dfrac{x^2+y^2}{x+y} \ge \dfrac{2xy}{\sqrt{2}} = \sqrt{2}xy$

as desired