Verified answer

We obviously have that:

0 ≤ (a - b)²

0 ≤ a² - 2ab + b²

a² + 2ab + b² ≤ 2a² + 2b² = 2c²

(a + b)² ≤ 2c²

We're done! :-)

EDIT: I should have just written:

Obviously,

(a + b)² ≤ (a - b)² + (a + b)² = 2c²

It would have been a nice one liner lol

EDIT: I think someone is trying to make it look like I'm the one who has given everyone a thumbs down because I'm the only answerer who doesn't have one.

@DUKE: That's a really nice proof :-)

EDIT: WOW, the most thumbs up I've ever got! :-)

EDIT: I'm curious, what was the purpose of this question? Because I'm sure you could have proved it yourself, so there must be something behind this...

it is impossible to factorize RHS to have a component that has LHS. Because you would ultimately want to prove that RHS = LHS + X where X ≥ 0. However you can prove it by another way which is to completely expand RHS and compare it with the expanded LHS to prove a similar point. Which is RHS = LHS + X where X ≥ 0. But it is not classy and very tedious. So for lazy ppl, like me, we would want to prove it by simply looking and comparing it. Since a b c are +ve real numbers (inclusive of 0), Let c ≥ b ≥ a ≥ 0 be true, then as u can see, the smallest value a,b or c can hold is 0. hence as the person below proved it, (0^5 - 0^2 +3)^3 ≥ (0 + 0 +0 )^3 => 27 is more than 0 (true statement) a^5 ≥ a^2 IFF a takes on a ≥ 1, therefore we have to analyse 2 scenarios for a ≥ 1 and 1 > a ≥ 0 (fractions) for a ≥ 1, a^5 ≥ a^2 thus, a^5-a^2 would always be ≥ 0 and minimum value of RHS is 27 for 1 > a ≥ 0 a^2 ≥ a^5 and a^5-a^2 < 3, infact it would always be < -1. therefore, a^5 - a^2 + 3 would always be +ve (it would approach a number bt for this qn we won't need to noe wad tat number is; ploting the graph and the minimum of x +ve would be it) since, the above would apply for both the b and the c components of the RHS, RHS is always +ve when a, b, c are +ve real numbers. and now for the LHS, when a,b,c are ≥ 1, RHS ≤ LHS as RHS has the additional 3 as well as the higher power. when 1 > c ≥ b ≥ a ≥ 0 (they are all fractions): when a approaches 0 from 1, (a^5 - a^2 + 3) would be U-shaped graph from 0 to 1, where the minimum is approximately 2.6743 when a approximately 0.7368. this would be the same for b and c. Hence, RHS is always ≥ LHS

In this case, the heavy hand of calculus does not provide a short

or elegant proof, but I always like seeing more than one approach.

Show that maximum value for a + b = c sqrt(2).

Proof:

a² + b² = c²

a² = c² - b²

a = sqrt(c² - b²)

a + b = sqrt(c² - b²) + b

Derivative of which is (sqrt(c² - b²) - b) / sqrt(c² - b²)

Setting that to 0, we have

sqrt(c²-b²) = b

c² - b² = b² = a²

c² = 2b² (** see note)

b = a = c/sqrt(2) = sqrt(2) c/2

At the maximum, a + b = sqrt(2) c,

a + b <= sqrt(2) c.

QED.

(** note) We can also reach this point by seeing

that the derivative is also equal to (a -b) / a

Setting a - b = 0, we have a = b,

a² + b² = 2b² = c².

.

Proof

(a-b)^2 = a^2+b^2-2ab ≥ 0

c^2 = a^2+b^2 ≥ 2ab

(a+b)^2 = a^2+b^2+2ab = c^2+2ab ≤ 2c^2, since c^2 ≥ 2ab

Take square root,

a+b ≤ c√2

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

Geometric meaning of the inequality:

In any right triangle, if the sum of two legs is a constant, then the minimum of hypotenuse is obtained iif the two legs are congruent.

a,b,c sides of a right triangle.

Since a, b and c are positive real numbers:

a+b > V2 c <=> (a+b)**2 >2c**2 <=> 2ab > c**2.

This is geometrically impossible (see proof below).

Another way to see it:

Let A be one of the acute angles.

Then

a+b > V2 c <=>

2 c**2 cos A sin A = c**2 sin (2A) > c**2 <=> sin (2A) > 1 which again is impossible.

Ana

EDIT

Sorry I didnt type my proof before but I wasnt at home and I typed my reply using my cellphone...

Sorry about this terrible drawing...

.... ...... .. C'

........ ......|. ...... C

........ a ...| ....... | .. b

...... ..... ..|... .... |

B ----- --- | -- ---- | ----- A

........c... H'..... H

C belongs to a circle since ACB is a right angle.

Area of ABC = ab/2 = ch/2, where h is HC

BTW, C' is the triangle whose area is maximum.

h <= c/2

ab/2 = ch/2 <= c^2/4

Hence 2 ab <= c^2

Ana

a= c cos t

and b = c sin t

so a + b = c ( cos t + sin t)

as c is constant we need to maximise

cos t + sin t

this is of the form A cos t + B sin t

A = 1 and B =1

let A = 1 =r cos a

B = 1 = r sin a

square and add r = sqrt(2) and tan a = 1

so 1 cos t + 1 sin t

= r cos a cos t + r sin a sin t

= r cos(t-a)

= sqrt(2) cos(t-pi/4)

a+b = sqrt(2) cos(t-pi/4) c

so <= sqrt(2) C as cos(t-pi/4) <= 1

(a-b)^2 >= 0 since a square is alway > 0

a^2 - 2ab + b^2 >= 0

a^2 + b^2 >= 2ab

2a^2 + 2b^2 >= a^2 + 2ab + b^2 (add a^2 + b^2 to both sides)

2c^2 >= (a+b)^2 (using a^2 + b^2 = c^2)

sqrt(2) c >= a+b since a,b,c>0

One of my favorite questions, thanks Dr D, I can hardly refrain from answering, even coming too late!

Fix c and follow the link below:

http://farm4.static.flickr.com/3115/2890530634_21a...

here is my answer

http://img401.imageshack.us/img401/533/c2a2b2nt2.j...

I hope this helps