HELP!

[neon grotto]

solarunes

[short orchid]

the second one

can we prove that

$\newline\frac{a+2b}{4b}\ge\frac{2a}{a+2b}\newline$

[neon grotto]

ILYAS

[short orchid]

just by knowing that a and b are strictly positive real numbers
???

hmm ic

[neon grotto]

solarunes

[short orchid]

wait

what does that mean ^

oww allr

I saw that something was missing

thx

anyway

.close

[violet stone]

Solved

Post marked as solved by @short orchid.

Use .unsolved if this was a mistake.

[neon grotto]

solarunes

[short orchid]

no prob wait

nah its not totally clear for me im gonna ask my teacher tomorow

ye thats it

im gonna try now

why not

allr

thx

i have to go for now

[lament epoch]

by cross multiply you get $(a+2b)^2 \geq 8ab \Leftrightarrow a^2+4ab+4b^2 \qeq 8ab \Leftrightarrow a^2+4b^2 \geq 4ab$ and from that, if you put $4ab$ to the LHS becomes $a^2-4ab+4b^2=(a-2b)^2 \geq 0$ and that hold for $\forall a,b \in \mathbb{R}$

[neon grotto]

Guinny42
Compile Error! Click the reaction for more information.
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[short orchid]

.sloved

[short orchid]

@urban shore in the end the solution was to subtract the two terms

😭

[urban shore]

Yeah, as the other guy pointed out. Guess I made it over complicated.