In-e-qualities

Calculus solution :

We only need to minimise $a^{2}+b^{2}$ since $\sqrt{x}$ is a monotonically increasing function.

We first express $b$ in terms of $a$ : $b = -\frac{3}{4}a + \frac{15}{4}$ . Then we get that $a^{2}+b^{2} = a^{2}+(-\frac{3}{4}a + \frac{15}{4})^{2}$ . To minimise this, we differentiate wrt $a$ and set to $0$ :

$\frac{d}{da} \left[ a^{2}+(-\frac{3}{4}a + \frac{15}{4})^{2} \right] = 2a+2(-\frac{3}{4}a + \frac{15}{4})(-\frac{3}{4}) = \frac{25}{8}a-\frac{45}{8} = 0$

(Here, we employed some assistance from the chain rule.)

Solving, we get $a = \frac{9}{5}$ . Now, plugging it all back into the original expression, we have

$\sqrt{a^{2}+b^{2}} = \sqrt{a^{2}+(-\frac{3}{4}a + \frac{15}{4})^{2}} = \sqrt{(\frac{9}{5})^{2}+(-\frac{3}{4} \frac{9}{5} + \frac{15}{4})^{2}} = \sqrt{9} = \boxed{3}$

Super nasty, though.

Geometric solution :

Notice the question is asking for the shortest distance between the origin and some point $(a,b)$ on the linear graph $3a+4b = 15$ . Of course, that is when the distance between the origin and the line is perpendicular!

Here, I've marked out the distance in blue. How do we find it, then? We could do a tedious calculation, solving for $a$ and $b$ , of course; however, see that a triangle is formed, and we can use its area to get the distance.

Let our distance be $x$ . Then

$\text{area of} \ \triangle = \frac{1}{2} \times 5 \times \frac{15}{4} = \frac{75}{8} = \frac{1}{2} \times \sqrt{5^{2}+(\frac{15}{4})^{2}}x = \frac{25x}{8}$

It's clear, then, that $x = \boxed{3}$ .

I thought that when adding squares that have bases that add to a constant, the smallest sum is when the bases are equal. So I set a=b and solved based on that. This did not get me the right answer, though it was close (within a couple hundredths). Can anyone explain why this is wrong?

My way is trial and error, there's only 2 ways it could've worked. a=5 and b=0 or a=1 and b=3 and √25 is more than √10 so √10 is the obvious choice

Is there any way we can solve this problem using the AM GM inequality?

There is a very nice solution to these kind of problems, we were taught in high school in Hungary, even in weaker ones. Look at this slick solution with vectors.

Look at the vectors (a,b) and (3,4). Take their scalar product.

One way to write it is $scalar=|a,b| |3,4| cos(gamma)=5cos(gamma)\sqrt{a^2+b^2}$

On the other hand, it equals: $scalar=x 1x 2+y 1y 2=3a+4b=15$ We get that $15=5cos(gamma)\sqrt{a^2+b^2}$ $3=cos(gamma)\sqrt{a^2+b^2}$. Since cos(gamma) is less or equal to 1, $\sqrt{a^2+b^2}$ is more or equal to 3.

Equality case: When gamma=0°.