Prove $a^4+b^4≥a^3b+ab^3$ for any $(a,b)\in\mathbb R$

Question

I need to prove $a^4+b^4≥a^3b+ab^3$ for any $(a,b)\in\mathbb R$.

What I did :-

$a^4+b^4≥a^3b+ab^3\Rightarrow a^4+b^4-a^3b-ab^3≥0\Rightarrow (a^3-b^3)(a-b)≥0\Rightarrow(a-b)^2(a^2+ab+b^2)≥0$

$(a-b)^2≥0$ can be easily proved but I am stuck at proving $(a^2+ab+b^2)≥0$.

Can anyone help?? Any other method is warmly welcomed.

Answer 1

You have

$$a^2+ab+b^2 = (a+b/2) ^2 -b^2/4 + b^2=(a+b/2) ^2 + 3b^2/4 \ge 0$$

Answer 2

Alternatively using AM-GM inequality, we have $$ \frac{a+b}{2} \ge \sqrt{ab}$$ or $$ (a+b)^2 \ge 4ab \ge ab $$ so that $$ (a+b)^2 -ab = a^2 + b^2 + ab \ge 0$$

Answer 3

$$a^4+b^4\ge a^3b+ab^3 \implies a^3(a-b)-b^3(a-b) \ge 0 \implies (a^3-b^3)(a-b) \ge 0.$$ The last part is obviously true.

Answer 4

Alternatively we can write,

$$\begin{align}&a^2+ab+b^2=k\\ \implies &a^2+ab+(b^2-k)=0\\ \implies &\Delta_a=b^2-4(b^2-k)≥0\\ \implies &\Delta_a=4k-3b^2≥0\\ \implies &k≥\frac 34b^2≥0.\end{align}$$

Answer 5

Prove $a^4+b^4≥a^3b+ab^3$ for any $(a,b)\in\mathbb R$

Alternative approach:

Without loss of generality, $0 \leq a,b$. This follows, since the LHS is unaffected by whether either or both of $a,b$ is negative. Further, the RHS is the same if $a,b$ are both positive or both negative. Also, the RHS becomes smaller if exactly one of $a,b$ is negative.

Since $(a-b)^2 \geq 0$, you have that

$$a^2 + b^2 \geq 2ab.\tag1 $$

Relying on the fact that $0 \leq a,b$, multiply both sides of (1) above by $(a \times b)$.

This gives

$$a^3b + ab^3 \geq 2a^2b^2.$$

Multiplying this by $3$ gives

$$3a^3b + 3ab^3 \geq 6a^2b^2.\tag2 $$

You also know that $(a - b)^4 \geq 0.$

Therefore

$$a^4 + b^4 + 6a^2b^2 \geq 4a^3b + 4ab^3.\tag3 $$

Combining (2) and (3) above yields

$$a^4 + b^4 + 3a^3b + 3ab^3 \geq 4a^3b + 4ab^3.$$

This implies that

$$a^4 + b^4 \geq a^3b + ab^3.$$

Answer 6

I feel like you've gotten a number of useful but complex answers. I'm going to go as basic as possible, which ends up meaning a proof by exhaustion. First, remember that all (real) squares are positive. Let $N = a^2 + ab + b^2$. Here are the cases:

$$a \geq b > 0 \implies N > 0 \tag{1}$$

$$a \leq b < 0 \implies N > 0 \tag{2}$$

These are pretty basic; if both numbers are positive, or both are negative, then all three terms in $N$ are positive. We should also check the cases with $0$s:

$$a = b = 0 \implies N = 0 \tag{3}$$

$$a \neq 0, b = 0 \implies N = a^2 > 0 \tag{4}$$

For the rest of the cases, one of the numbers is negative; we'll choose $b < 0$. We can substitute $-b$ for $b$ everywhere, and use $N' = a^2 - ab + b^2$ instead. This means, then, that $N' \geq 0$ is equivalent to $a^2 + b^2 \geq ab$.

$$a = b \implies a^2 + a^2 > a^2 \ (= ab) \tag{5}$$

$$a > b \implies a^2 > ab \implies a^2 + b^2 > ab + b^2 > ab \tag{6}$$

This exhausts all the cases, and in each case either $N \geq 0$ or $N' \geq 0$. Hence $a^2 + ab +b^2 \geq 0$ for all real numbers.