Find the minimum value of $a^4+b^4+\sqrt {ab}$ if $a+b=1$. Find the minimum value of $a^4+b^4+\sqrt {ab}$ if $a+b=1$.

My attempt:
If we put $a=b=\dfrac {1}{2}$, we get $\dfrac {5}{8}$ which is the minimum value. But are there any other methods.?
Note that: 

$$a+b=1 \Rightarrow a^2+b^2=1-2ab \Rightarrow a^4+b^4=1+2a^2b^2-4ab.$$

If we denote $\sqrt{ab}=t$, the LHS function will be:

$$f(t)=2t^4-4t^2+t+1.$$

Noting that $0\le \sqrt{ab} \le 0.5,$ minimum will be $f(0.5)=0.625$.
To prove that the minimum of the function is at $\frac{5}{8}$, let us assume that we have a positive real number $k$, that is the absolute value of difference between $a$ or $b$ and $\frac{1}{2}$. Any other pairs of solutions can be represented as $(0.5+k)^4 + (0.5-k)^4 + \sqrt{(0.5+k)(0.5-k)}$. 

Using the binomial theorem, we have:
$$(0.5+k)^4 = [(1)(0.5)^4 (k)^0 + (4)(0.5)^3(k)^1 + (6)(0.5)^2(k)^2 + (1)(0.5)^1 (k)^3$$

$$(0.5-k)^4 = [(1)(0.5)^4 (k)^0 + (4)(0.5)^3(-k)^1 + (6)(0.5)^2(-k)^2 + (1)(0.5)^1 (-k)^3 + (4)(0.5)^0(-k)^4]$$

The terms that have $1$st, $3$rd powers cancel, leaving:

$$(0.5+k)^4+(0.5-k)^4 = [(2)(0.5)^4 (k)^0+ (12)(0.5)^2(k)^2 + (2)(0.5)^0(-k)^4]$$
$$=2*\frac{1}{16} + 12*\frac{1}{4}*k^2 + 2*k^4$$
$$=2k^4+3k^2+\frac{1}{8}$$

The last term $\sqrt{(0.5+k)(0.5-k)}$ can be simplified into $\sqrt{-k^2+0.25}$ by the difference of squares.

Adding the terms together gives:
$$=(2k^4+3k^2+0.125) + (\sqrt{-k^2+0.25})$$

Since the derivative of $2k^4+3k^2+0.125$ is larger than $-k^2+0.25$, it follows that $2k^4+3k^2+0.125$ is larger than $\sqrt{-k^2+0.25}$, so the first function grows faster. 

The first term has a minimum of $0.125$ when $k=0$, and also when $k=0$ the function $\sqrt{-k^2+0.25}$ evaluates to $0.5$, $0.5 + 0.125 = 0.625$, so the function is at is minimum when $a,b $ = $0.5+0$, or when $a,b = 0.5$.