# If $a>0$, $b>0$, $\frac{1}{a}+\frac{8}{b}=1$, what is the minimum of $a^2+b^2$?

I change $$\frac{1}{a}+\frac{8}{b}=1$$ into $$\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2}=1$$, then I get $$a^2+b^2=(a^2+b^2)(\frac{1}{a^2}+\frac{16}{ab}+\frac{64}{b^2})=65+\frac{16a}{b}+\frac{16b}{a}+\frac{64a^2}{b^2}+\frac{b^2}{a^2}$$

though I get $$\frac{16a}{b}+\frac{16b}{a}\geq 32$$,$$\frac{64a^2}{b^2}+\frac{b^2}{a^2}\geq 16$$, their equivalency condition is not identical.

How can I solve it?

• Holders Inequality will yield least value as $125$ when $a=5,b=10$ Apr 6, 2023 at 4:25

Solution: The minimum value is $$125$$ at $$a = 5$$ and $$b = 10$$.

Approach: First let's express $$b$$ in terms of $$a$$. From $$\frac{1}{a} + \frac{8}{b} = 1$$, we have

$$1 - \frac{1}{a} = \frac{8}{b} \implies \frac{a - 1}{8a} = \frac{1}{b}$$ $$\implies b = \frac{8a}{a-1}$$

and now since $$b$$ must be $$> 0$$, we have $$\frac{8a}{a - 1} > 0 \implies a > 1$$ which is stronger than $$a > 0$$ (from problem statement).

Thus, all that is left to find is the minimum of the equation $$a^2 + b^2$$ or, $$f(a) = a^2 + \frac{64 a^2}{(a - 1)^2}$$ where $$a>1$$. This minimum (which happens to be 125) can be easily found via differentiation.

Differentiation: The required minima will be obtained when $$\frac{df(a)}{da} = 0$$ where $$a > 0$$.

$$\frac{df(a)}{da} = 0,\ \text{where } a > 0$$ $$\implies \frac{2a(a-5)(a(a + 2) + 13)}{(a-1)^3} = 0,\ \text{where } a > 0$$ $$\implies (a - 5) > 0 \implies a = 5$$ $$\implies a = 5,\ b= \frac{8\times 5}{4} = 10$$

Thus, the minimum is obtained at $$a = 5, b = 10$$ and the value is $$a^2 + b^2 = 25 + 100 = 125$$ as needed.

Note that I have skipped some steps (when coming to $$a - 5 > 0$$) and arguments (prove that the point $$a = 5$$ is indeed a minima for $$f(a)$$ using $$f''(a)$$ at $$a = 5$$) here.

• Your right, nice solution Apr 6, 2023 at 4:19
• @NoChance for $a = 3$ and $a = 9$, $\frac{df(a)}{da}$ is not zero right? They are not stationary points. So you won't find a global (or local) minima at those two points.
– Arjo
Apr 6, 2023 at 5:18
• You are correct. Thanks. Apr 6, 2023 at 5:29

Hint: Using Hölder's inequality $$\left(a^2+b^2 \right)\left(\frac1a+\frac8b \right)^2\geqslant (1+4)^3$$ with equality possible when $$a^2:\dfrac1a=b^2:\dfrac8b$$

• wow! that's right! Apr 6, 2023 at 4:50
• How do you use Holder's inequality here? Do you take $p = q = 2\$? What is your $x_1,x_2,y_1$ and $y_2\$? What I know is that if $\frac {1} {p} + \frac {1} {q} = 1$ then Holder's inequality says that $\sum\limits \left \lvert x_i y_i \right \rvert \leq \left (\sum\limits |x_i|^p \right )^{\frac {1} {p}} \left (\sum\limits |y_i|^q \right )^{\frac {1} {q}}.$ Apr 6, 2023 at 5:04
• Oh! I got it. You take $p = 3$ and $q = \frac {3} {2}$ and $x_1 = a^{\frac {2} {3}}, x_2 = b^{\frac {2} {3}}, y_1 = \frac {1} {x_1}$ and $y_2 = \frac {4} {x_2}.$ Apr 6, 2023 at 5:33
• @AnilBagchi. This is a simpler version of the same inequality, essentially as you noted, $p=3, q = \frac32$ in the original form. Apr 6, 2023 at 5:37

Another approach is to let $$a = x+1, b=8(y+1)$$ which gives $$\frac 1{x + 1} + \frac{1}{y + 1} =1$$

which is just $$y = 1/x$$. This can be parameterized by $$(x, y) = p(t) = (e^{-t}, e^{t})$$.

And the original constraints $$a > 0, b > 0$$ implies $$a > 1, b > 8$$ which is equivalent to $$x > 0, y > 0$$ which is implied by the parameterization.

Then it becomes minimizing $$f(t) = (e^{-t} + 1)^2 + 64(e^t + 1)^2$$ which is straightforward to work out, gives the same result as previously posted answers.