145 Followers

I wanted to put emphasis on a common misuse of the AM-GM inequality on $1 + a^4 + b^8$ :

Since a, b are positive, $1 + a^4 + b^8 \geq 3 \times \sqrt[3]{a^4 b^8} = 3(a b^2)^{\frac{4}{3}} = 3 \times 5^{\frac{4}{3}}$

Equality occurs when $1 = a^4 = b^8 = 5^{\frac{4}{3}}$ . But $1 = 5^{\frac{4}{3}}$ is a contradiction. So while $3 \times 5^{\frac{4}{3}}$ represents a lower bound on the expression, it does $\underline{not}$ a minimum.

Instead, apply the AM-GM Inequality on $a^4 + b^8$ .

$a^4 + b^8 \geq 2 \times \sqrt[2]{a^4 b^8}= (a b^2)^{2} = 2 \times 5^2$

Equality occurs when $a^4 = b^8 = 25 \Rightarrow a = \sqrt[2]{5}, b = \sqrt[4]{5}$ .

Therefore, $min(a^4 + b^8) = 50 \Rightarrow min(1 + a^4 + b^8) = 1+ 50 = \boxed{51}$ (which I should note is greater than $3 \times 5^{\frac{4}{3}}$ ) :)

$b^2=\dfrac 5 a. ~~~~~~~~~~~~~~~ b^8=\left(\dfrac 5 a \right)^4.\\ f(a)=1+a^4+\left(\dfrac 5 a \right)^4, ~~ f '(a)=0 \implies ~~ a=\sqrt 5.\\ f "(a)>0, ~~ so ~ ~~~ a=\sqrt 5. ~ is ~ min.\\ f(a)_{min}=1+(\sqrt 5)^4+\left(\dfrac 5 {\sqrt 5} \right)^4=\Large \color{#D61F06}{51}$

a and b being positive integers does not satisfy the answer . They must be positive reals.

Seems extremely awkward the number 145. You should have waited for 150 or maybe you should have infact written 150 as no one is there who will check your profile before attempting the question.