# Math Help - Prove that

1. ## Prove that

2. Originally Posted by dhiab
Let b = 1 - a.
Gives:
$\sqrt { a^2 + 1 - a} + \sqrt{ 1 - a + a^2} + \sqrt{ 1 + a - a^2}$

0 <= a <= 1
so the max value that any of the three square roots can have is 1.
(you may wish to elaborate on this)

And the sum of the three square roots must then be less than or equal to 3.

3. Originally Posted by Unenlightened
0 <= a <= 1
so the max value that any of the three square roots can have is 1.
(you may wish to elaborate on this)
The first two square roots are $\le1$ but the last one isn’t.

$0\le a\le1\ \implies\ \sqrt{1+a-a^2}\ge1$

4. Ah feck. In future, I shall be more careful...

Edit: In the words of H.H.Munro
"A little inaccuracy sometimes saves a world of explanation."

Right so, the third square root may be greater than 1, but for every increase above 1 in that, there is double the decrease in the sum of the other two square roots, so the greater the value of the third square root, the lower the value overall.
I doubt that counts as a proof though..

5. By the Generealised-Mean Inequality $(*)$: $
\left( {\tfrac{{\sqrt {a^2 + b} + \sqrt {b^2 + a} + \sqrt {1 + ab} }}
{3}} \right)^2 \leqslant \left( {\tfrac{{\left( {a^2 + b} \right) + \left( {b^2 + a} \right) + \left( {1 + ab} \right)}}
{3}} \right)^1
$

Now: $
\left( {a^2 + b} \right) + \left( {b^2 + a} \right) + \left( {1 + ab} \right) = \left( {a + b} \right)^2 + \left( {a + b} \right) + 1 - ab = 3 - ab \leqslant 3
$

Thus: $
\left( {\tfrac{{\sqrt {a^2 + b} + \sqrt {b^2 + a} + \sqrt {1 + ab} }}
{3}} \right)^2 \leqslant 1
$
from which we get: $
\sqrt {a^2 + b} + \sqrt {b^2 + a} + \sqrt {1 + ab} \leqslant 3
$
$\square$

$(*)$ See here

6. I think Cauchy–Schwarz might also work here.

Note that the sum of the terms inside the square roots is $a^2+ab+b^2+a+b+1=(a+b)^2+a+b+1-ab=3-ab\leq3.$

$\therefore\ 1\cdot\sqrt{a^2+b}+1\cdot\sqrt{a+b^2}+1\cdot\sqrt{ 1+ab}$

$\le\ \sqrt{(1^2+1^2+1^2)(3-ab)}$

$\le\ \sqrt{3\cdot3}$

$=\ 3$