Find Min of $(a-1)^4+(b+2)^4$ w/ $a+b\ge 3$

[anemone]

Let $a$ and $b$ be real numbers such that $a+b\ge 3$.

What is the minimum value of the expression $(a-1)^4+(b+2)^4$?

 

[kaliprasad]

Let $a$ and $b$ be real numbers such that $a+b\ge 3$.

What is the minimum value of the expression $(a-1)^4+(b+2)^4$?

Spoiler

let $m = a - 1$ and $n = b+ 2$
we get $m+n = a + b + 1 >= 3+ 1 >= 4$
we need to minimize $m^4+n^4$ given $m+n>=4$ and from law of symmetry it is minumum at
$m=n=2$ or $a=3, b=0$ and value is $32$. as at a = 1 b = 2 it is $4^4= 256$ larger

 

[MarkFL]

Spoiler

let $m = a - 1$ and $n = b+ 2$
we get $m+n = a + b + 1 >= 3+ 1 >= 4$
we need to minimize $m^4+n^4$ given $m+n>=4$ and from law of symmetry it is minumum at
$m=n=2$ or $a=3, b=0$ and value is $32$. as at a = 1 b = 2 it is $4^4= 256$ larger

Spoiler

I like how you cleverly coaxed cyclic symmetry from the problem...well played! (Yes)

 

[Greg]

Let $a$ and $b$ be real numbers such that $a+b\ge 3$.

What is the minimum value of the expression $(a-1)^4+(b+2)^4$?

Spoiler

The lowest values for $a$ and $b$ must occur when $a+b=3$. This can be seen by graphing the inequality $y\ge3-x$ on the Cartesian axes. As the exponents in the expression are even, negative numbers do not effectively reduce the sum.

Lagrange multipliers:

$$\Lambda=(a-1)^4+(b+2)^4-\lambda(a+b-3)$$
$$\dfrac{d\Lambda}{da}=4(a-1)^3-\lambda=0$$
$$\dfrac{d\Lambda}{db}=4(b+2)^3-\lambda=0$$
$$\dfrac{d\Lambda}{d\lambda}=a+b-3=0$$
$$\Rightarrow4(a-1)^3=4(b+2)^3$$
$$\implies a-1=b+2$$
$$\Rightarrow a-b=3\Leftrightarrow a+b=3$$
$$\Rightarrow2a=6\Rightarrow a=3,\,b=0$$
$$\min\left[(a-1)^4+(b+2)^4\right]=2^4+2^4=32$$

 

[anemone]

Thanks all for participating!(Cool)

My solution:

Spoiler

\(\displaystyle \begin{align*}(a-1)^4+(b+2)^4&\ge \frac{((a-1)^2+(b+2)^2)^2}{1+1}\text{by the extended Cauchy-Schwarz inequality}\\&\ge \frac{\left(\frac{(a-1+b+2)^2}{1+1}\right)^2}{1+1}\text{again by the extended Cauchy-Schwarz inequality}\\&= \frac{\left(\frac{(a+b+1)^2}{2}\right)^2}{2}\\& \ge \frac{\left(\frac{(3+1)^2}{2}\right)}{2}\text{since}\,\,\,a+b\ge 3\\& =32\end{align*}\)

Equality occurs when $a=3,\,b=0$.