# Prove the inequality $(ab + 1)(a + b) \geq 4ab$ for any non-negative real numbers $a$ and $b$. When does equality hold?

Prove the inequality

$$(ab + 1)(a + b) \geq 4ab$$

for any non-negative real numbers $$a$$ and $$b$$. When does equality hold?

Attempt:

To prove the inequality

$$(ab + 1)(a + b) \geq 4ab$$

for any non-negative real numbers $$a$$ and $$b$$, we start by expanding the left-hand side:

$$(ab + 1)(a + b) = a^2b + ab^2 + a + b.$$

This gives us the inequality:

$$a^2b + ab^2 + a + b \geq 4ab.$$

Rearranging the terms, we get:

$$a^2b + ab^2 + a + b - 4ab \geq 0.$$

Next, we apply the AM-GM inequality, which states that:

$$a + b \geq 2\sqrt{ab}.$$

Substituting this into our expression, we have:

$$(ab + 1)(a + b) \geq (ab + 1) \cdot 2\sqrt{ab}.$$

Simplifying further, we get:

$$2ab\sqrt{ab} + 2\sqrt{ab} \geq 4ab.$$

Dividing both sides by $$2\sqrt{ab}$$, we find:

$$ab + 1 \geq 2\sqrt{ab}.$$

To check the equality condition, we set:

$$ab + 1 = 2\sqrt{ab}.$$

Squaring both sides, we obtain:

$$(ab - 1)^2 = 0 \implies ab = 1.$$

In conclusion, the inequality $$(ab + 1)(a + b) \geq 4ab$$ holds for all non-negative $$a$$ and $$b$$, with equality if and only if $$ab = 1$$. Is this ok? Thanks for your assistence.

• It looks to me as if your demonstration may be upside down: it is conventional to start which what you know is true and finish with what you want to prove. Commented Jun 19 at 23:51
• You could have AM-GM directly (termwise). IE $(ab+1)(a+b) \geq 2\sqrt{ab} \times 2 \sqrt{ab} = 4 ab$. $\quad$ However, determining the equality case is a bit trickier, because you have to be careful about multiplying by 0. Commented Jun 20 at 21:53

Notice that $$(ab+1)(a+b)-4ab = (a-1)^2b+(b-1)^2a \ge 0$$

Equality occurs only if $$(a-1)^2b = 0$$ and $$(b-1)^2a = 0$$, which has solutions $$(a,b)=(0,0)$$ and $$(a,b)=(1,1)$$.

If $$a=b=0$$ or if $$a=0, b\ne 0$$ then the inequality holds.

Now assume that $$a,b\ne 0$$ let $$ab =x^2$$ where $$x> 0$$

By AM-GM inequality $$a+b \ge 2 \sqrt{ab}=2x$$ $$(ab+1)(a+b)-4ab \ge 2x(x^2+1)-4x^2 = 2x(x^2+1-2x)=2x(x-1)^2\ge0$$ So $$(ab+1)(a+b)-4ab \ge 0$$

$$(ab+1)(a+b)-4ab=0$$ hold when $$a=b=0$$.

Assume $$a,b>0$$

Since $$(ab+1)(a+b)-4ab \ge 2x(x-1)^2$$, $$x=1$$ should hold so the equality holds

Now lets assume $$ab=1$$ then $$b=\frac{1}{a}$$

$$2(a+\frac 1 a)-4 =0$$ $$a+\frac 1 a =2$$ $$a^2-2a +1 =0$$ $$a=1$$

So $$a=b=1, a=b=0$$ are the only solutions for the equality

You could have AM-GM directly (termwise).

$$(ab+1)(a+b) \geq 2\sqrt{ab} \times 2 \sqrt{ab} = 4 ab.$$

To determine the equality cases, that can happen when:

• $$ab+1 = 2 \sqrt{ab} = 0$$ : No solutions.
• $$a+b = 2\sqrt{ab} = 0$$: $$(a,b) = (0,0)$$.
• $$ab+1 = 2\sqrt{ab}, a+b = 2\sqrt{ab}$$: $$(a, b) = (1, 1)$$.