Inequality: Prove that sqrt(x+y)<= sqrt(x) + sqrt(y) for x,y >= 0

[jeszo]

Homework Statement

Prove that √x+y ≤ √x + √y for all x,y ≥ 0

Homework Equations

The Attempt at a Solution

square both sides: x + y ≤ x + 2√x√y + y

subtracting x and y: 0 ≤ 2√x√y

dividing by 2: 0 ≤ √x√y

0 ≤ √x√y is true for all x,y since the square root of a number is always non negative, and two non negatives multiplied together gives you a non negative number.

----------------------------

I submitted this proof and got 2/10 on it, but I have no clue where I went wrong. Am I missing something obvious?

 

[susskind_leon]

Well, you probably should have done the steps the other way around, starting with 0 ≤ √x√y and arriving at √x+y ≤ √x + √y

 

[CAF123]

I think you have the chain of causality the wrong way round. Your attempt has assumed the correctness of the statement already. Try proving by contradiction, that is, assume $$\sqrt{x+y} > \sqrt{x} + \sqrt{y}\,\,\,\,\forall\,x,y\,\in ℝ_{≥0}$$ and see if this leads to an absurdity.

 

[jeszo]

Thanks for your responses, I understand my mistake now

 

[madah12]

wait why is it wrong again? I thought if you start with an assumption but go with <=> and reach Tautology then the assumption is true? like prove x+1>x
x+1>x if and only if 1>0 which is true so x+1>x
so why couldn't he say
sqrt(x) + sqrt(y) >= sqrt(x+y) given that x,y >0
if and only if x+y+2sqrt(xy)>x+y
if and only if 2sqrt(xy)>0
<=>sqrt(xy)>0
<=> sqrt(x) > 0 and sqrt(y)>0
<=>x,y >0
which is true due to given constraint?
would this constitute a proof ?
wait why is it wrong again? I thought if you start with an assumption but go with <=> and reach Tautology then the assumption is true? like prove x+1>x
x+1>x if and only if 1>0 which is true so x+1>x
so why couldn't he say
sqrt(x) + sqrt(y) >= sqrt(x+y) given that x,y >0
if and only if x+y+2sqrt(xy)>x+y
if and only if 2sqrt(xy)>0
<=>sqrt(xy)>0
<=> sqrt(x) > 0 and sqrt(y)>0
<=>x,y >0
which is true due to given constraint?

Also what about this method of solving it?
(i can give a solution since the solver already did it right?)
y= x+c such that x+c>=0 infinity>c>=-x
then we look at sqrt(x) + sqrt(x+c) >= sqrt(2x+c)
when x =0 or x=-c
sqrt(x) + sqrt(x+c) = sqrt(2x+c)
when x=!0 and x!=-c
f(x)= sqrt(x) + sqrt(x+c)-sqrt(2x+c)
f(0)=0
f'(x)= 1/2sqrt(x) + 1/2sqrt(x+c) -1/4sqrt(2x+c)
f'(x) >0
if
1+ sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0
1+ sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)> sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)
if sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0 then sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)+1>0
sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0 => 1/x+c >1/4(2x+c) => 1/y >1/4(x+y) ,since x>0 ,1/y>1/(x+y)>1/4(x+y)
=>sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0, 1+sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c) >sqrt(x)/sqrt(x+c) -sqrt(x)/2sqrt(2x+c)>0
=>f'(x)>0
f(0)>0 , f'(x)>0 for all x >0 so f(x)>0 for all x >0
sqrt(x) + sqrt(x+c)-sqrt(2x+c)>0
sqrt(x) + sqrt(x+c)>sqrt(2x+c)
sqrt(x) +sqrt(y)>sqrt(x+y)

I mean for me intuitively it makes sense and that was all that was needed in my calculus course ;o

 

[CAF123]

wait why is it wrong again? I thought if you start with an assumption but go with <=> and reach Tautology then the assumption is true? like prove x+1>x
x+1>x if and only if 1>0 which is true so x+1>x
so why couldn't he say
sqrt(x) + sqrt(y) >= sqrt(x+y) given that x,y >0
if and only if x+y+2sqrt(xy)>x+y
if and only if 2sqrt(xy)>0
<=>sqrt(xy)>0
<=> sqrt(x) > 0 and sqrt(y)>0
<=>x,y >0
which is true due to given constraint?
would this constitute a proof ?

Yes, I believe so. The $<=>$ symbol is key.