forgot what to cycle (2 Viewers)

ExtremelyBoredUser · Mar 25, 2022

Run hard@thehsc said:
View attachment 35274

$(\sqrt{a} - \sqrt{b})^2 \geq 0$
$a + b \geq 2\sqrt{ab}$
$\frac{a+b}{2} \geq \sqrt{ab}$ (x)

Let a = p + q/2 , b = r + s/2
$\frac{p+q+r+s}{4} \geq \sqrt{\frac{p+q}{2} \times \frac{r+s}{2}}$
$\frac{p+q+r+s}{4} \geq \sqrt{\sqrt{pqrs}}$ as from (x)

hence $\frac{p+q+r+s}{4} \geq (pqrs)^{\frac{1}{4}}$

let s = a + b + c/3, p = a, q = b, r = c
$\frac{4(a+b+c)}{3 \times 4} \geq (\frac{abc(a+b+c)}{3})^{\frac{1}{4}}$

$(\frac{(a+b+c)}{3})^{4} \geq (\frac{abc(a+b+c)}{3})$

$(\frac{(a+b+c)}{3})^{3} \geq (abc)$

Hence

$(\frac{(a+b+c)}{3}) \geq (abc)^{\frac{1}{3}}$

There's other ways of doing this but this is the one I remember off the head

Run hard@thehsc · Mar 25, 2022

as from (x)
- @ExtremelyBoredUser could you pls explain this step - did not get this....

ExtremelyBoredUser · Mar 25, 2022

Run hard@thehsc said:
as from (x)
- @ExtremelyBoredUser could you pls explain this step - did not get this....

in (x) $\frac{a+b}{2} \geq \sqrt{ab}$ hence $\frac{p+q}{2} \geq \sqrt{pq}$ and similarly for r and s

This is just simplifying the RHS, so you get

$\sqrt{\frac{p+q}{2} \times \frac{r+s}{2}} \geq \sqrt{\sqrt{pq} \times \sqrt{rs}}$

and hence the LHS ;

$\frac{p+q+r+s}{4} \geq \sqrt{\frac{p+q}{2} \times \frac{r+s}{2}} \geq \sqrt{\sqrt{pqrs}}$

forgot what to cycle (2 Viewers)

Run hard@thehsc

Well-Known Member

ExtremelyBoredUser

Bored Uni Student

Run hard@thehsc

Well-Known Member

ExtremelyBoredUser

Bored Uni Student

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