AM-GM?

Algebra Level 3

Positive reals a a and b b are such that a + b = s a+b=s . The maximum value of a + b s \dfrac{\sqrt a+\sqrt b}{\sqrt s} is c \sqrt c . Find c c .


The answer is 2.00.

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7 solutions

By Cauchy-Schwarz inequality , we have:

a + b s ( 1 + 1 ) ( a + b ) s = 2 s s = 2 \begin{aligned} \frac {\color{#3D99F6} \sqrt a + \sqrt b}{\sqrt s} & \le \frac {\color{#3D99F6}\sqrt{(1+1)(a+b)}}{\sqrt s} = \frac {\sqrt{2s}}{\sqrt s} = \sqrt 2 \end{aligned}

Therefore, c = 2 c=\boxed 2 .

@Alak Bhattacharya , for square root you can enter \sqrt 2 2 \sqrt 2 , \sqrt {2\pi} 2 π \sqrt{2\pi} , \sqrt[3]8 = 2 8 3 = 2 \sqrt[3] 8 = 2 . Put all formulas a + b = s a+b=s , all variables x , y , z x, y, z and constants a , b , c a, b, c in LaTex.

Chew-Seong Cheong - 1 year, 11 months ago

Smart substitution 2:

Given (√a+√b)/√(a+b)=√c

Write a=s(1+x)/2 b=s(1−x)/2

So √c=[√(1+x)+√(1−x)]/√2

Then c=1+1√(1-x²)

Whence √c=√[1+1√(1-x²)]

Thus √c≤√2

Max.c=2

Shailesh Patel - 1 year, 6 months ago
Chris Lewis
Jul 1, 2019

Here's a somewhat inelegant solution.

Without loss of generality, we can set s = 1 s=1 (this just scales a a and b b by the same factor and doesn't affect the final result). So we have b = 1 a b=1-a , and we want to maximise f ( a ) = a + 1 a f(a)=\sqrt{a}+\sqrt{1-a} , where 0 a 1 0 \le a \le 1 .

We can use calculus for this, but it's neater to observe that both a \sqrt{a} and 1 a \sqrt{1-a} are concave functions, so their sum is, too. Also, f ( a ) f(a) is symmetric about a = 1 2 a=\frac12 ; these two facts mean its maximum value can only occur at a = 0 a=0 (and a = 1 a=1 ), or at a = 1 2 a=\frac12 . It's easy to check that the largest of these is when a = 1 2 a=\frac12 , giving c = 2 c=2 .

Alternatively, if we say a = x 2 a=x^2 and b = y 2 b=y^2 , this problem becomes "maximise x + y x+y subject to x 2 + y 2 = 1 x^2+y^2=1 ", which is a slightly more familiar formulation and can be solved in a number of different ways (for instance, geometrically - plotting in the x y x-y plane, the constraint is a circle and the function to be maximised a straight line).

Emanuele Prati
Jul 1, 2019

We know that a + b s c \frac{\sqrt{a}+\sqrt{b}}{\sqrt{s}}\le \sqrt{c} . Squaring both sides of the equation (they are both positive) we get a + b + 2 a b s c \frac{a+b+2\sqrt{ab}}{s}\le c . From AM-GM (geometric mean is always less or equal the arithmetic one) we know that a b a + b 2 \sqrt{ab}\le \frac{a+b}{2} which means 2 a b a + b = s 2\sqrt{ab}\le a+b=s . Therefore a + b + 2 a b s a + b + s s = 2 s s = 2 c \frac{a+b+2\sqrt{ab}}{s}\le \frac{a+b+s}{s}=\frac{2s}{s}=2 \le c , so c = 2 c=\boxed{2} (because c c is the maximum and this expression cannot be greater than 2).

Aareyan Manzoor
Jul 1, 2019

this problem is equivalent to minimizing a + b r \dfrac{a+b}{r} for the eqn a 2 + b 2 = r 2 a^2+b^2=r^2 .

Geometrically, this is the equivalent of finding the line tangent to a circle with the eqn a + b = k a+b= k , i.e a 4 5 45^\circ line.

this is an isosceles right triangle, meaning we would have r 2 + r 2 r \dfrac{\sqrt{r^2+r^2}}{r} be the minimum, i.e 2 \boxed{\sqrt{2}}

Aaghaz Mahajan
Jul 1, 2019

Hint :

Use the power mean inequality, i.e.

a 1 p + a 2 p + . . . + a n p n p a 1 q + a 2 q + . . . + a n q n q \sqrt[p]{\frac{a_1^p+a_2^p+...\ +\ a_n^p}{n}}\ge\sqrt[q]{\frac{a_1^q+a_2^q+...\ +\ a_n^q}{n}}

where, a i \displaystyle a_i is a positive real number and p > q \displaystyle p>q

No restriction over p and q? Will the inequality hold irrespective of whether p is less or more than q?

A Former Brilliant Member - 1 year, 11 months ago

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Ohh srry I forgot to mention that p > q p>q

Aaghaz Mahajan - 1 year, 11 months ago

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Now its Okay.

A Former Brilliant Member - 1 year, 11 months ago
Jesse Nieminen
Jul 1, 2019

Substitute x = a x = \sqrt{a} and y = b y = \sqrt{b} to the original expression yielding x + y x 2 + y 2 \dfrac{x + y}{\sqrt{x^2 + y^2}} .

Now by the QM-AM inequality x + y x 2 + y 2 = x + y 2 2 x 2 + y 2 2 x 2 + y 2 2 2 x 2 + y 2 2 = 2 , \dfrac{x + y}{\sqrt{x^2 + y^2}} = \dfrac{x + y}{2} \sqrt{\dfrac{2}{x^2 + y^2}} \sqrt{2} \leq \sqrt{\dfrac{x^2 + y^2}{2}} \sqrt{\dfrac{2}{x^2 + y^2}}\sqrt{2} = \sqrt{2},

where equality is attained when x = y x = y .

Hence c = 2 c = \boxed{2} .

Kris Hauchecorne
Jul 2, 2019

a = u²

b = v²

s = a + b = u² + v² = r²

u = r cos(t)

v = r sin(t)

y = cos(t) + sin(t)

y' = -sin(t) + cos(t) = 0

t = pi/4 en y = sqrt(2)

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