# Prove that ${ a }^{ 2 }+2ab+{ b }^{ 2}\ge 0$ without using $(a+b) ^{ 2 }$

Prove that $${ a }^{ 2 }+2ab+{ b }^{ 2 }\ge 0,\quad\text{for all }a,b\in \mathbb R$$ without using $(a+b)^{2}$.

My teacher challenged me to solve this question from any where. He said you can't solve it. I hope you can help me to solve it.

• Use $AM>=GM$ inequality – Archis Welankar Jan 25 '16 at 15:25
• Take partial derivatives, alternatively. – MathematicsStudent1122 Jan 25 '16 at 15:30
• Can you use $(a-b)^2 \geq 0$ ? – lisyarus Jan 25 '16 at 15:31
• @ArchisWelankar AM-GM inequality applies to nonnegative numbers, no? – leonbloy Jan 25 '16 at 15:31
• @leonbloy WLOG suppose $a>0,$ $b<0$ and apply AM-GM to $a$ and $-b.$ – Justpassingby Jan 25 '16 at 15:42

## 8 Answers

Alternatively, you can study the familly of functions

$$f_b(x) = x^2 +2bx + b^2$$

It's easy to see that the minimum of this function is attained at $x = -b$, so the minimum of $f_b$ is $0$ for all values of parameters $b$, hence $\forall a,b, \quad a^2+2ab+b^2 \geq 0$

Define a function $$f(a,b)=a^2+2ab+b^2$$

And try to find its extrema:

$$\begin{cases} f'_a=0\\ f'_b=0 \end{cases}$$

$$\begin{cases} 2a+2b=0\\ 2b+2a=0 \end{cases}$$

$$\begin{cases} a+b=0 \end{cases}$$

Since $f''_{ab}{}^2-f''_{aa} f''_{bb}=4-4=0$ there may be an extremum or not. Try to compute $f(a,-a-1)$, $f(a,-a)$ and $f(a,-a+1)$. You will see that

$$f(a,-a-1)=f(a,-a+1)=1\quad\text{but}\quad f(a,-a)=0$$

It means that this function has infinitely many weak minima that lay on the line $a+b=0$. It follows that

$$f(a,b)\geqslant 0$$

since $f$ is continuous (and smooth) as a polynomial.

This is more of an intuitive explanation:

The statement is clear when $a$ and $b$ have the same sign, because the left hand side is strictly positive. This reduces to showing

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

for $a,b > 0$.

Think in terms of money. Assume you have an $a$-valued coin, and a $b$-valued coin. Assume further that $a \geq b$. Say you are allowed to take $x > 0$ of one coin and $y > 0$ of another coin. How would you maximize your profit? Of course you take most of the coin with the largest value, and least of the one with lowest value. This would be greater than taking the least of the coin with the greatest value and the most of the coin with the lowest value. Now take $x = a$ and $y = b$.

Case 1. If both $a,b$ are positive , then each term $a^2, 2ab, b^2$, all are positive. Hence $a^2 + 2ab +b^2>0$.

Case 2. If both $a,b$ are negative, then also each term $a^2, 2ab, b^2$, all are positive. Hence $a^2 + 2ab +b^2>0$.

Case 3. If one of $a,b$ is positive and other is negative, then in this case by changing one of its sign sufficient to show $a^2 -2ab+b^2 > 0$, where $a,b$ both are positive. So, without loss of generality assume $a>b$. Here is a pictorial proof of that fact. Without loss of generality suppose $0<-b<a$ and write $c=-b.$

Multiplying a positive number by the positive scale factor $r=\frac a c>1$ has more effect on large positive numbers than on small positive numbers. Thus

$$ra-a>rc-c$$

in other words

$$a^2/c-a>a-c$$

multiplying by $c$ yields

$$a^2-ac>ac-c^2$$

or

$$a^2+2ab+b^2>0$$

(the inequality is strict because we ruled out the trivial case $c=a$ up front)

Express the pair $(a,b) \in \mathbb{R}^2$ using polar coordinates

$$a=r \cos \theta,b=r \sin \theta .$$

We have

$$a^2+2ab+b^2=r^2+2r^2 \cos \theta \sin \theta =r^2(1+\sin 2 \theta) \geq 0.$$

Using the rearrangement inequality on two instances of the sequence $\{a,b\}$

$ab + ba \le a^2 + b^2$

Intuitive reasoning, if you're multiplying two sequences of numbers pairwise, the maximal value is when the sequences are ordered the same way. Simplifying further, assume both sequences are positive numbers, take one sequence fixed and pick multipliers from the second sequence as weights. You'll achieve the maximum value if you multiple the largest number with a the largest weight and so on.

If $a,b$ have the same sign then all summands are non-negative. So, let $a>0$ and $b<0$ (without loss) and write $$a^2+2ab+b^2=a^2-2a|b|+b^2=a^2-a|b|+|b|^2-a|b|=a(a-|b|)+|b|(|b|-a)$$ Now, take two more cases

1. if $a\ge |b|$, then $a(a-|b|)+|b|(|b|-a)\ge a(a-|b|)+a(|b|-a)=0$ and similarly
2. if $a<|b|$, then $a(a-|b|)+|b|(|b|-a)>|b|(a-|b|)+|b|(|b|-a)=0$.