A generic inequality

Notice that $a+b+c=1$ :

$\displaystyle a^4+b^4+c^4\\ =\dfrac{1}{2}(a^4+b^4+a^4+c^4+b^4+c^4)\\ \geq\dfrac{1}{2}(2a^2b^2+2a^2c^2+2b^2c^2)\\ =\dfrac{1}{2}(a^2b^2+a^2c^2+a^2b^2+b^2c^2+a^2c^2+b^2c^2)\\ \geq\dfrac{1}{2}(2ab\cdot ac+2ab\cdot bc+2ac\cdot bc)\\ =a^2bc+ab^2c+abc^2\\ =abc(a+b+c)\\ =abc$

The two equalities are achieved if and only if $a=b=c=\dfrac{1}{3}$ , so $a^4+b^4+c^4$ can be equal to $abc$ , thus the geq sign.

Almost the same as Alice 's solution.

We know that, since $a, b, c$ are reals, therefore

$(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\geq 0\implies a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$ .

Now, $a^2b^2+b^2c^2\geq 2ab^2c$ (A. M. - G. M. inequality)

$b^2c^2+c^2a^2\geq 2abc^2$

and $c^2a^2+a^2b^2\geq 2a^2bc$ .

Therefore $2(a^2b^2+b^2c^2+c^2a^2)\geq 2abc(a+b+c)\implies a^2b^2+b^2c^2+c^2a^2\geq abc$ . (since $a+b+c=1$ ).

Hence, $a^4+b^4+c^4\geq abc$ .

Without any loss of generality, assume $a\geq b \geq c.$ We have $a^4 + b^4 +c^4 = a^2\cdot a^2 + b^2 \cdot b^2 + c^2 \cdot c^2 \stackrel{(a)}{\geq} a^2 bc + b^2 ca + c^2 ab = abc (a+b+c)\stackrel{(b)}{=} abc.$ where the step (a) follows from the Rearrangement inequality , and the step (b) follows from the fact that $a+b+c=1.$

I just put: $a = 1, b = 1 , c = 0$ and $a^4 + b^4 + c^4 = abc$ so I put yes.