# Thread: Proving inequality with a, b

1. ## Proving inequality with a, b

Prove that for any positive real numbers a, b this inequality is valid:

2. Assuming a and b are positive numbers, then
for the first inequality:
$\displaystyle \text{ since }2t^4-5t^3+6t^2-5t+2=(t-1)^2(2t^2-t+2)\geq 0 \text{ holds for all }t>0$.
$\displaystyle \text{ By letting }t=\sqrt{\frac{a}{b}},\text{ we have } 5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2$,
Both sides times $\displaystyle b^2$, gives the first inequality.

for the second inequality:
$\displaystyle \text{ since }a^2+b^2\geq 2ab,\text{ we have }$
$\displaystyle 4a^2+12ab+4b^2\leq 5a^2+10ab+5b^2=5(a+b)^2$.
Both sides divided by $\displaystyle 10(a+b)$ gives the second inequality.

3. Please, I don't understend how have you come to this inequality?

By the way, do you mean first inequality as this ineq.:: $\displaystyle \sqrt{ab}\leq \frac{2(a^2+3ab+b^2)}{5(a+b)}$

How have you found out that t is $\displaystyle \sqrt{\frac{a}{b}}$?

What is this inequality deduced from?$\displaystyle 5(\frac{a}{b}+1)\sqrt{\frac{a}{b}}\leq 2(\frac{a}{b})^2+6(\frac{a}{b})+2$?? .....
...

4. $\displaystyle \sqrt{ab}\leq\frac{2(a^2+3ab+b^2)}{5(a+b)}\leq\fra c{a+b}{2}$

$\displaystyle 2\sqrt{ab}\leq\frac{4(a^2+3ab+b^2)}{5(a+b)}\leq\ a+b$

right inequality..

$\displaystyle \frac{4(a+b)^2+4ab}{5(a+b)}\leq\ a+b$

$\displaystyle 4(a+b)^2+4ab\leq\ 5(a+b)^2$

$\displaystyle 4ab\leq\ (a+b)^2$

$\displaystyle (2\sqrt{ab})^2\leq\ (a+b)^2$

$\displaystyle 2\sqrt{ab}\leq\ a+b$

$\displaystyle a+b-2\sqrt{ab}\geq0$

Since $\displaystyle (\sqrt{a}-\sqrt{b})^2\geq0$

then $\displaystyle a+b-2\sqrt{ab}\geq0$

Next prove the left inequality...

5. Big thanx.

6. Left inequality...

$\displaystyle 5\sqrt{ab}\leq\frac{2(a+b)^2+2ab}{a+b}$

$\displaystyle 5\sqrt{ab}\leq2(a+b)+\frac{2ab}{a+b}$

We know from the right inequality that $\displaystyle a+b\geq2\sqrt{ab}$

Hence $\displaystyle 2(a+b)\geq4\sqrt{ab}$

We are now asking if $\displaystyle \frac{2ab}{a+b}\geq\sqrt{ab}$

$\displaystyle \frac{2\sqrt{ab}\sqrt{ab}}{a+b}\geq\sqrt{ab}\ ?$

$\displaystyle 2\sqrt{ab}\geq(a+b)\ ?$ No..

inconclusive

We can say... $\displaystyle 2(a+b)\ is\ \geq4\sqrt{ab}$
by 2 times the amount that
$\displaystyle a+b\ is\ \geq2\sqrt{ab}.$

We can examine the case for $\displaystyle \sqrt{ab}>\frac{2\sqrt{ab}\sqrt{ab}}{a+b}$

Difference is

$\displaystyle (\sqrt{ab})\frac{a+b-2\sqrt{ab}}{a+b}$

the ratio is

$\displaystyle \frac{\sqrt{ab}}{a+b}(a+b-2\sqrt{ab})$

Since... $\displaystyle a+b\ge2\sqrt{ab}$

then.... $\displaystyle 1\geq\frac{\sqrt{ab}}{a+b}$

The ratio is $\displaystyle \leq1$,

therefore $\displaystyle \sqrt{ab}\geq\frac{2ab}{a+b}$ by a margin not greater than

the difference between $\displaystyle 2(a+b)$ and $\displaystyle 4\sqrt{ab}$