MHB Inequality Challenge: Prove $b^3+a^3 \le 2$

[anemone]

The positive real $a$ and $b$ satisfy $b^3+a^2\ge b^4+a^3$. Prove that $b^3+a^3\le 2$.

 

[Opalg]

It ought to be easier than this!
[sp][graph]iubck9bicm[/graph] (Click on the diagram for an enlargement.)

This is based on the idea that, within the positive quadrant, the region enclosed by the brown curve (1) is contained in the interior of the green circle (2), which in turn is contained in the the region enclosed by the blue curve (3). These regions are given by the inequalities $x\geqslant0$, $y\geqslant0$, together with $$(1)\quad x^2 - x^3 + y^3 - y^4 \geqslant0,$$ $$(2)\quad x+y-x^2 - y^2 \geqslant0,$$ $$(3)\quad 2 - x^3 - y^3 \geqslant0.$$ To show that (1) implies (2), we need to show that $x+y-x^2 - y^2 \geqslant x^2 - x^3 + y^3 - y^4.$ That is equivalent to $x-2x^2 + x^3 + y - 2y^2 + y^3 = x(1-x)^2 + y(1-y)^2 \geqslant0$, which is obviously true when $x\geqslant0$ and $y\geqslant0.$

I thought that it ought to be easier to prove that (2) implies (3), because the green and blue curves have a more regular shape than the brown one. But the only way I can see to prove $(2)\Rightarrow(3)$ is to use Lagrange multipliers. In fact, to minimise the distance from a point on the blue curve to the centre of the green circle, we have to minimise $f(x,y) = \bigl(x-\frac12\bigr)^2 + \bigl(y-\frac12\bigr)^2$ subject to the condition $x^3+y^3 = 2$. To do that, put the partial derivatives of $f(x,y) - \lambda(x^3+y^3-2)$ equal to zero. That gives $$2x+1 - 3\lambda x^2 = 0,\qquad 2y-1 - 3\lambda y^2 = 0,$$ from which $\dfrac{2x-1}{x^2} = \dfrac{2y-1}{y^2}$. That reduces to $(x-y)(2xy-x-y) = 0$. You can check that the only point on the curve $x^3+y^3 = 2$ satisfying either of the conditions $x-y=0$ or $2xy-x-y=0$ is the point $(1,1)$. Therefore the blue curve lies entirely outside the green circle except at that one point. That is equivalent to the implication $(2)\Rightarrow(3)$.

Since $(1)\Rightarrow(2)$ and $(2)\Rightarrow(3)$, it follows that $(1)\Rightarrow(3)$. In particular, if you put $(x,y) = (a,b)$, it follows that $b^3+a^2\geqslant b^4+a^3$ implies $b^3+a^3\leqslant 2$.[/sp]

 

[anemone]

It ought to be easier than this!
[sp][graph]iubck9bicm[/graph] (Click on the diagram for an enlargement.)

This is based on the idea that, within the positive quadrant, the region enclosed by the brown curve (1) is contained in the interior of the green circle (2), which in turn is contained in the the region enclosed by the blue curve (3). These regions are given by the inequalities $x\geqslant0$, $y\geqslant0$, together with $$(1)\quad x^2 - x^3 + y^3 - y^4 \geqslant0,$$ $$(2)\quad x+y-x^2 - y^2 \geqslant0,$$ $$(3)\quad 2 - x^3 - y^3 \geqslant0.$$ To show that (1) implies (2), we need to show that $x+y-x^2 - y^2 \geqslant x^2 - x^3 + y^3 - y^4.$ That is equivalent to $x-2x^2 + x^3 + y - 2y^2 + y^3 = x(1-x)^2 + y(1-y)^2 \geqslant0$, which is obviously true when $x\geqslant0$ and $y\geqslant0.$

I thought that it ought to be easier to prove that (2) implies (3), because the green and blue curves have a more regular shape than the brown one. But the only way I can see to prove $(2)\Rightarrow(3)$ is to use Lagrange multipliers. In fact, to minimise the distance from a point on the blue curve to the centre of the green circle, we have to minimise $f(x,y) = \bigl(x-\frac12\bigr)^2 + \bigl(y-\frac12\bigr)^2$ subject to the condition $x^3+y^3 = 2$. To do that, put the partial derivatives of $f(x,y) - \lambda(x^3+y^3-2)$ equal to zero. That gives $$2x+1 - 3\lambda x^2 = 0,\qquad 2y-1 - 3\lambda y^2 = 0,$$ from which $\dfrac{2x-1}{x^2} = \dfrac{2y-1}{y^2}$. That reduces to $(x-y)(2xy-x-y) = 0$. You can check that the only point on the curve $x^3+y^3 = 2$ satisfying either of the conditions $x-y=0$ or $2xy-x-y=0$ is the point $(1,1)$. Therefore the blue curve lies entirely outside the green circle except at that one point. That is equivalent to the implication $(2)\Rightarrow(3)$.

Since $(1)\Rightarrow(2)$ and $(2)\Rightarrow(3)$, it follows that $(1)\Rightarrow(3)$. In particular, if you put $(x,y) = (a,b)$, it follows that $b^3+a^2\geqslant b^4+a^3$ implies $b^3+a^3\leqslant 2$.[/sp]

Thanks for participating, Opalg! This is a very good solution that is so well explained. I salute you for this.:)

Solution proposed by other:

Spoiler

This is equivalent in asking us to show that if $b^3+a^3>2$, then we must have $b^3+a^2<b^4+a^3$.

Power-Mean inequality tells us

$\sqrt[3]{\dfrac{b^3+a^3}{2}}\ge \sqrt[2]{\dfrac{b^2+a^2}{2}}$

Rearrange and manipulate the inequality above we get

$\left( \dfrac{b^3+a^3}{2} \right)^{\dfrac{2}{3}}(2) \ge b^2+a^2$

$(b^3+a^3)^{\dfrac{2}{3}}(2)^{\dfrac{1}{3}} \ge b^2+a^2$

Since $b^3+a^3>2$, we can say that

$(b^3+a^3)^{\dfrac{2}{3}}(b^3+a^3)^{\dfrac{1}{3}} \ge b^2+a^2$

$b^3+a^3 \ge b^2+a^2$

$\therefore b^3-b^2 \ge a^2-a^3$

On the other hand, note that $b^2(b-1)^2 \ge 0$, this gives $b^4-b^3 \ge b^3-b^2$.

Hence we get

$b^4-b^3 \ge a^2-a^3$

$b^4+a^3\ge b^3+a^2$

$b^3+a^2<b^4+a^3$, proving the case.