# Prove that $a^4+b^4+1\ge a+b$.

Prove that $$a^4+b^4+1\ge a+b$$ for all real numbers $$a,b$$.

What I've tried:

1.I checked how AM-GM may help but doesn't look like it's useful here.

1. I've tried:

$$(a^2+b^2)^2 -2(ab)^2+1 \ge a+b$$

But unfortunately, I can't find any way to continue this.. I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.

*This exercise is from the TAU entry exams.

# Proof

Since $$\left(a^4-a^2+\frac{1}{4}\right)+\left(b^4-b^2+\frac{1}{4}\right)=\left(a^2-\frac{1}{2}\right)^2+\left(b^2-\frac{1}{2}\right)^2 \geq0,$$ hence $$a^4+b^4+1 \geq a^2+b^2+\frac{1}{2}.\tag1$$

Since $$\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)=\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2 \geq 0,$$ hence, $$a^2+b^2+\frac{1}{2} \geq a+b.\tag2$$

Combining $$(1)$$ and $$(2)$$, $$a^4+b^4+1 \geq a+b.$$

• Can you please elaborate a bit more? I'm having trouble to understand the logic between each step.
– user565804
Jul 21 '18 at 12:26
• this is a purely elementary proof，which just brings in an intermediate variable. Can you agree that $x^2-2xy+y^2=(x-y)^2 \geq 0$? Jul 21 '18 at 14:30
• Yes of course I can see that, It's just that it's hard for me to see how the first row of the proof leads to the second - that's the only thing.
– user565804
Jul 22 '18 at 8:06
• @Maxim Do you know the transposition of the inequality？ For example， $x-y \geq 0$ may imply $x \geq y$. Jul 22 '18 at 14:37
• ok, sure, that makes sense but the part that was strange for me is why the second step has a positive one half and not a negative, after looking at it one more time, I perfectly understand everything, indeed a very nice solution!
– user565804
Jul 22 '18 at 16:27

Note that, by the AM-GM Inequality, $$x^4+\frac{1}{2}=x^4+3\left(\frac{1}{6}\right)\geq 4\sqrt[4]{x^4\left(\frac{1}{6}\right)^3}=\sqrt[4]{\frac{256}{216}}\,|x|\geq |x|\geq x$$ for all $x\in\mathbb{R}$. The equality does not hold, though.

A sharper inequality is $a^4+b^4+\frac{3}{\sqrt[3]{2}^5}\geq a+b$. This is because $$x^4+\frac{3}{\sqrt[3]{2}^8} =x^4+3\left(\frac{1}{\sqrt[3]{2}^8}\right)\geq 4\sqrt[4]{x^4\left(\frac{1}{\sqrt[3]{2}^8}\right)^3}=|x|\geq x\,.$$ The inequality above is an equality if and only if $x=\frac{1}{\sqrt[3]{2}^2}$.

• How did you solve it so quickly? i.e. what ideas came to your mind when you read it? Jul 20 '18 at 20:35
• The first idea was to try to separate the variables. So, when I saw the problem statement, I immediately thought that the inequalities $a^4+\frac12\geq a$ and $b^4+\frac12\geq b$ should hold. Jul 20 '18 at 20:37
• Thanks; I will keep this tactic in mind. Jul 20 '18 at 20:38
• $$x^4 + 3(\frac{1}{6}) \geq 4(x^4(\frac{1}{6})^3)^{1/4}$$ How did you do that using $$\sum\limits_{i=1}^n a_i\lambda_i \geq \Pi a_i^{\lambda_i}$$ I could see that $\lambda_2 = 3$ and $\lambda_1 = 1$ But where did the 4th root come from ? Thanks Jul 20 '18 at 21:11
• $$x^4,\frac{1}{6},\frac{1}{6},\frac{1}{6}$$ are four numbers. Jul 20 '18 at 21:18

We have

$$a^4+b^4+1\ge a+b\iff a^4-a+b^4-b+1\ge 0$$

and it is easy to show that

$$f(x)=x^4-x\implies f'(x)=4x^3-1\implies x_{min}=\frac1{4^\frac13}\quad f(x)\ge f(x_{min})\approx-0.4725$$

• You could work with $f(x)=x^4-x+\frac 12$ to preserve the symmetry and since $f(x)>0$ then $f(a)+f(b)$ too. Why keep the $1$ apart ?
– zwim
Jul 20 '18 at 21:37
• @zwim Yes of course w ecan also do in that way, it is almost equivalent.
– user
Jul 20 '18 at 21:41

Clues only in this answer, since you said you want to figure out the answer yourself.

Have you tried proving it for positive numbers greater than $2$? The square of such a number is obviously greater than $2$, and much more so the biquadrate. Prove it for those numbers and the answer is obvious for negative numbers less than $-2$.

Where it gets really tricky is in the unit interval. Let's say $$a = b = \frac{1}{2}.$$ Then $$\left(\frac{1}{2}\right)^4 = \frac{1}{16},$$ so $$a^4 + b^4 = \frac{1}{8},$$ which is actually less than either $a$ or $b$ alone. But then you get to add $1$ to that...

Let's have $4c^3=1$ and $f(a)=a^4-a+\frac 12$

\begin{align}\require{cancel}f(c+u)-f(c) &=\cancel{c^4}+\cancel{4c^3u}+6c^2u^2+4cu^3+u^4-\cancel{c}-\cancel{u}+\cancel{\frac 12}-\cancel{c^4}+\cancel{c}-\cancel{\frac 12}\\ &=u^2\underbrace{(u^2+4cu+6c^2)}_{\Delta=-8c^2<0}\ge 0\end{align}

So $f(c)$ is a minimum and $f(a)\ge f(c)\approx0.0275>0$

The conclusion arises from $f(a)+f(b)>0$.