If $a,b> 0$, $a\neq b$, and $a + b = 2$ prove that $ab < 2$.

Question

Let $a$ and $b$ be two positive real numbers such that $a \neq b$.

Also, $a+b=2$. Now it is required to prove that $ab<2$. Thanks for any systematic and mathematical proof.

Answer 1

$(a+b)^2 =4 $ gives $ ab=2-\frac{a^2+b^2}{2}<2$

Answer 2

GM inequality $\frac{a+b}{2}\geq \sqrt{ab}$ thus $\frac{a+b}{2}=\frac{2}{2}\geq \sqrt{ab}$ implies $ab\leq 1<2$

https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

Answer 3

We have $(a-b)^2\ge 0$ and in fact $>0$ since we are given that $a\ne b$. Then $$ 4=(a+b)^2=(a-b)^2+4ab>4ab$$ and so $$ab<1.$$ (Note that this argument did not use $a,b>0$).

Answer 4

Hint.

If $a+b=2$ and $ab=c$ the two numbers are the solutions of $x^2-2x+c=0$.

Can you find whan this equation has two real positive solutions?

Answer 5

Ok, so $a,b \in \Bbb R_{> 0}$, $a \neq b$ and $a+b=2$.

Suppose $ab \geq 2$. Then $a \geq \frac{2}{b}$ since $b > 0$. Thus $\frac{2}{b}+b \leq 2 $ which leads to $b^2-2b+2 = (b-1)^2+1\leq 0$ and hence $(b-1)^2 \leq -1$, which is a contradiction. Thus $ab < 2$.

Answer 6

Hint: If a and b are positive numbers, and $a+b=2$, then $a=2\cos^2t$ and $b=2\sin^2t$ is a valid parameterization, wouldn't you agree ? Now use $\sin2t=2\sin t~\cos t$.