# 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.

• I don't see what's tedious about finding the extrema. How can $x^3=(1-x)^3$? Remember that $f(x)=x^3$ is a one-to-one function. – Ted Shifrin Jan 12 '17 at 21:46
• Well, I reckon used the wrong word. I just don't think it's the best method to prove such inequality. – VanDerWarden Jan 12 '17 at 21:47
• Sometimes, such inequalities have for example clear geometric interpretation. – VanDerWarden Jan 12 '17 at 21:48
• Extrema of symmetric functions occur symmetrically, but not always when $x=y$. But for such simple functions, it's always when $x=y$. You can find a "simpler" argument for $x^2+y^2$ and then just substitute. – Ted Shifrin Jan 12 '17 at 21:50

## 10 Answers

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.

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*}

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.

$x^4$ 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 $x=y$.

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.

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 $$k=\left(\frac 12\right)^2+\left(\frac 12\right)^2=\frac 18$$

$$(x+y)^4=1 \\ x^4+4x^3y+6x^2y^2+4x^3+y^4 =1$$

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 $$1=x^4+4x^3y+6x^2y^2+4x^3+y^4 \leq x^3+(3x^4+y^4)+3(x^4+y^4)+(x^4+3y^4)+y^4$$

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 $$x^t+y^t\le 2^{1-t}$$

use Holder inequality we have $$(x^4+y^4)(1+1)(1+1)(1+1)\ge (x+y)^4$$

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 $\frac{1}{8}$.