/sci/ - Science & Math

>>10494905
just do the

the thing with the numbers and lines

bam, result, QED

>>10494905
I mean, it's pretty obvious.
Both terms are greater than or equal to 0 when they're defined.
When x > 0, x + 3 > 3, so sqrt(x + 3) > sqrt(3). Then the whole thing is.
When x = 0, the fourth root of 9 is sqrt(3). So the sum is 2sqrt(3).
When x < 0, 9 - x > 9. Then the fourth root of 9 - x is greater than the fourth root of 9, which is sqrt(3). Then the sum of both terms is greater than sqrt(3).
Since we've shown it for negative x, positive x, and 0, we have shown it for all x.

>>10494905
Why don't you elevate the inequation to the forth power and be done with it

-3 <= x <= 9

>>10494922
>Since we've shown it for negative x, positive x, and 0, we have shown it for all x.
>pretty obvious
[math]x = -4[/math]
[math]\sqrt{-1}+\sqrt[4]{13}>\sqrt{3}\ \ ?[/math]

Photomath
>>10494905

>>10494905
https://www.google.com/search?q=y%3D%28x%2B3%29%5E.333%2Cy%3D%289-x%29%5E.25%2Cy%3D3%5E.5

>>10494905
between -3<x<9,
always

Just (x+3)^(1/2)+(9-x)^(1/4)>3^1/2
t. retard

>>10495150
He is right, he is just missing the conditions of existence for the two even roots: -3<x<9 is the solution

>>10494905
>Posting high school math
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