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

Question

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.

Answer 1

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 $(a,b)=(1,1)$.

Answer 2

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 $(ab+1)(a+b)-4ab \ge 0$

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

Answer 3

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} $: $(a, b) = (1, 1)$.