# If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$

I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)

Question: If $$a^2+b^2 \gt a+b$$ and $$a,b \gt 0$$ Prove that $$a^3+b^3 \gt a^2+b^2$$

My solution: Let $$a \geq b$$

From $$a^2+b^2 \gt a+b$$ we get $$a^2-a \gt b-b^2$$

Since $$a \geq b$$ we can get $$a^3-a^2 \gt b^2-b^3$$ $$\Rightarrow$$ $$a^3+b^3 \gt a^2+b^2$$

If this solution is incorrect, please explain why and attach the correct solution. Thank you.

• $a \geq b$ and $c \geq d$ don't imply that $ac \geq bd$. – Kavi Rama Murthy Sep 28 '18 at 6:41
• @KaviRamaMurthy it's true if a,b,c,d are positive. – abc... Sep 28 '18 at 6:42
• @abc... In the argument $c$ is $a^{2}-a$ and $d$ is $b^{2}-b$. We do not know that $c,d \geq 0$. – Kavi Rama Murthy Sep 28 '18 at 7:14
• @user 587054 In fact the answer by Zamarion also appears to be wrong.A correct proof has been given by Michael Rozenberg. – Kavi Rama Murthy Sep 28 '18 at 7:33

## 4 Answers

It is correct, but I would explain the "since $$a\ge b$$ then $$a^3 - a^2 > b^3 - b^2$$" step a bit more. \begin{align*}a^2 - a > b^2-b &\iff a(a^2-a) > a(b^2-b) \quad \text{(since a>0)} \\ &\iff a(a^2-a)>a(b^2-b)\ge b(b^2-b) \quad \text{(since b\le a)} \\ &\iff a^3-a^2 > b^3 -b^2 \\ &\iff a^3 + b^3 > a^2 + b^2\end{align*}

• Thank you for your help. – user587054 Sep 28 '18 at 6:57
• I wanted to show that the condition $a>0$ is necessary for this step (it wasn't clear for me at first glance). I know this is overkill, but I thought that this look like a homework question and so, OP might want to explain every step clearly. – Zamarion Sep 28 '18 at 7:19
• @Zamarion Why is $a(b^{2}-b) \geq b(b^{2}-b)$? We do not know that $b^{2}-b \geq 0$. I think the proof is incorrect. – Kavi Rama Murthy Sep 28 '18 at 7:27
• $a\ge b$ so $a\ge 1$ – Toni Mhax Sep 28 '18 at 8:50
• You must start with $a^2 - a > b-b^2$. You copied/quoted from the OP incorrectly. – farruhota Sep 28 '18 at 9:20

Also, we can make the following.

Since by the condition $$1>\frac{a+b}{a^2+b^2},$$ by C-S we obtain: $$a^3+b^3>\frac{(a^3+b^3)(a+b)}{a^2+b^2}\geq\frac{(a^2+b^2)^2}{a^2+b^2}=a^2+b^2.$$

Your proof is correct. Indeed: $$\begin{cases}a^2-a \gt b-b^2\\ a\ge b>0\end{cases} \Rightarrow \\ a(a^2-a)>b(b-b^2) \Rightarrow \\ a^3-a^2>b^2-b^3 \Rightarrow \\ a^3+b^3>a^2+b^2.$$

Alternative proof. Consider $$b=ax, x\ge 1$$. Then: $$a^2+b^2 \gt a+b \Rightarrow \\ a^2+a^2x^2>a+ax \stackrel{\text{divide by} \ a}{\Rightarrow} \\ a+ax^2>1+x \Rightarrow \\ a>\frac{1+x}{1+x^2} \ \ (1)$$ Hence: $$a^3+b^3>a^2+b^2 \iff \\ a^3+a^3x^3>a^2+a^2x^2 \Rightarrow \\ a+ax^3>1+x^2 \iff \\ a(1+x^3)\stackrel{(1)}{>}\frac{1+x}{1+x^2}(1+x^3)\ge 1+x^2 \iff \\ (1+x)(1+x^3)\ge(1+x^2)^2 \iff \\ 1+x+x^3+x^4\ge1+2x^2+x^4 \iff \\ x(1+x^2)\ge 2x^2 \iff \\ (x-1)^2\ge 0.$$ It is the Cauchy-Swarz inequality (given by Michael Rozenberg).

I think the step when you arrive at $$a^3-a^2>b^2-b^3$$ is incorrect as you multiply inequalities whose expressions can be negative.

Alternatively, note that $$a^3+a\ge 2a^2$$ (this is equivalent to $$a(a-1)^2 \ge 0$$) and analogously $$b^3+b\ge 2b^2$$. Therefore $$a^3+b^3+a+b\ge 2(a^2+b^2) > a^2+b^2+a+b,$$ where the second inequality uses the assumption $$a^2+b^2>a+b$$. As a consequence, $$a^3+b^3>a^2+b^2$$.