For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
 A: It is equivalent to showing that $(1/2+\epsilon)^4+(1/2-\epsilon)^4\geq 1/8$. But this simplifies to $\tfrac{1}{8}(16\epsilon^4+12\epsilon^2+1)$, which obviously is minimized at $\epsilon=0$, giving $\tfrac{1}{8}$.
Edit: This is motivated by guessing the minimum, i.e. $x=y=1/2$, and then symmetrizing. The symmetry allows a nice expansion using the binomial theorem.
A: Also a Cauchy-Schwarz approach works:
\begin{align*}
1 = x + y \le \sqrt{(x^2+y^2)(1^2+1^2)} \implies x^2 + y^2 \ge \frac{1}{2} \\
\frac{1}{2} \le x^2 + y^2 \le \sqrt{(x^4 + y^4)(1^2 + 1^2)} \implies x^4 + y^4 \ge \frac{1}{8}
\end{align*}
A: If $a,b\in \mathbb{R}$, then $(a+b)^2\leq 2(a^2+b^2)$ since $2ab\leq a^2+b^2$.
Applying this with $x$ and $y$ we get
$$ 1=(x+y)^2\leq 2(x^2+y^2)$$
so $x^2+y^2\geq \frac{1}{2}$, and then applying the inequality again with $x^2$ and $y^2$ we get
$$ \frac{1}{4}\leq (x^2+y^2)^2\leq 2(x^4+y^4)$$
which is the desired result.
A: $x^4$ and $y^4$ are convex functions of $x$ and $y$, so their sum is also convex. The restriction of this to the line $x+y=1$ is again convex.  By symmetry, the minimum must occur at the point where $x=y$.
A: hint: I posted a solution to this same question a year ago, and I think you can search it here. You can use the inequality $\dfrac{a^2+b^2}{2} \ge \left(\dfrac{a+b}{2}\right)^2$ twice. 
A: If you are looking for a more geometrical explanation, consider the family of curves $$x^4+y^4=k$$
These are a family of concentric rounded square shaped curves with order 4 symmetry centred at the origin. The line $y=x$ is a line of symmetry. We need to find the value of $k$ for which such a curve is tangent to the line $x+y=1$.
By symmetry $x=y=\frac 12$ and hence $$k=\left(\frac 12\right)^2+\left(\frac 12\right)^2=\frac 18$$
A: $$(x+y)^4=1 \\
x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$
Now, by AM-GM we have
$$x^3y \leq \frac{x^4+x^4+x^4+y^4}{4}\\
x^2y^2 \leq \frac{x^4+y^4}{2}\\
xy^3 \leq \frac{x^4+y^4+y^4+y^4}{4}\\
$$
Thus
$$1=x^4+4x^3y+6x^2y^2+4x^3+y^4  \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$
A: If $x\le0$, then $y\ge1$, so $x^4+y^4\ge1$. Hence we can assume $x,y>0$.
For $x,y>0$, $x\ne y$, the function
$$
\mu(x,y;t)=\begin{cases}
\left(\dfrac{x^t+y^t}{2}\right)^{1/t} & \text{if $t\ne0$} \\[6px]
\sqrt{xy\vphantom{X}} & \text{if $t=0$}
\end{cases}
$$
is continuous and increasing in the variable $t$. Thus
$$
\mu(x,y;1)<\mu(x,y;4)
$$
that is
$$
\frac{x+y}{2}<\left(\dfrac{x^4+y^4}{2}\right)^{1/4}
$$
and, if $x+y=1$,
$$
\frac{1}{16}<\frac{x^4+y^4}{2}
$$
For $x=y$, the inequality is obvious (and is an equality, actually).

The proof that $\mu(x,y;t)$ is increasing (for $x\ne y$) is an application of convexity. Suppose $0<p<q$; we want to prove that
$$
\left(\dfrac{x^p+y^p}{2}\right)^{1/p}<
\left(\dfrac{x^q+y^q}{2}\right)^{1/q}
$$
that is,
$$
\left(\dfrac{x^p+y^p}{2}\right)^{q/p}<\dfrac{x^q+y^q}{2}
$$
Set $u=x^p$ and $v=y^p$; then the inequality becomes
$$
\left(\dfrac{u+v}{2}\right)^{q/p}<\dfrac{u^{q/p}+v^{q/p}}{2}
$$
which is a consequence of $z\mapsto z^{p/q}$ being convex (on $(0,\infty)$).
Since it's immediate that $\mu(x,y;t)$ is continuous at $0$, we have that it is increasing over $[0,\infty)$.
Now notice that
$$
\mu(x,y;-t)=\mu(x^{-1},y^{-1};t)^{-1}
$$
and we can conclude that the function is also increasing over $(-\infty,0]$.
This is a “generalized AM-GM” inequality, which is the simple observation that $\mu(x,y;0)<\mu(x,y;1)$. For $t=-1$ we get the harmonic mean.
The inequalities become nonstrict if and only if $x=y$.
More generally, then, for $x+y=1$, $x,y>0$ and $t>1$, we have $\mu(x,y;1)\le\mu(x,y;t)$, that is
$$
x^t+y^t\ge\frac{1}{2^{t-1}}
$$
If instead $0<t<1$, we have $\mu(x,y;t)\le\mu(x,y;1)$, so
$$
x^t+y^t\le 2^{1-t}
$$
A: use Holder inequality we have
$$(x^4+y^4)(1+1)(1+1)(1+1)\ge (x+y)^4$$
A: Apply Lagrangian multiplier method ! \begin{align}
\min \quad x^4 + y^4 \\ \text{s.t.} \quad x + y = 1 \end{align} gives $\mathcal{L}(x,y) = x^4 + y^4 - \lambda(x+y-1)$. FOC gives \begin{align} \mathcal{L}_x = 4x^3 - \lambda & = 0 \\ \mathcal{L}_y = 4y^3 - \lambda & = 0 \\ x+y & = 1. \end{align} Solving gives an optimum of $(x^*,y^*) = (\frac{1}{2},\frac{1}{2})$ with value $\frac{1}{8}$.
