$(x+y)^r < x^r + y^r$ whenever $x$ and $y$ are positive real numbers and $r$ is a real number with $0 < r < 1$.

In the solution it says it is safe to assume that $x+y=1$. I don't see any reason why this is the case... Why is it safe to assume $x+y=1$? If so how does this help proving this statement?


  • $\begingroup$ Removed the (proof-theory) tag since it did not seem relevant. $\endgroup$ – Srivatsan Aug 1 '11 at 16:02
  • $\begingroup$ Added the inequality tag. Not convinced the discrete-mathematics tag is appropriate. $\endgroup$ – Gerry Myerson Aug 2 '11 at 3:45

It is safe because you can divide both sides of the inequality by $(x+y)^r$, then substitute $x'=\frac{x}{x+y}, y'=\frac{y}{x+y}$ and have the same inequality with $x'+y'=1$. It looks like it helps by making the LHS be $1$, but without seeing the proof it is hard to comment further on how it helps.

  • $\begingroup$ @Mayumi BTW in many places, such a step (divide by $(x+y)^r$, so that the $x+y$ term can be assumed to be $1$) will be called "normalization". For e.g., the book could also say "By suitable normalization, it is safe to assume that $x+y=1$". $\endgroup$ – Srivatsan Aug 1 '11 at 15:54
  • 1
    $\begingroup$ It helps because once the reduction to $x+y=1$ is done, one has just to check that the one-variable function $u(x)=x^r+(1-x)^r$ is such that $u(x)<1$ for every $x$ in $(0,1)$--and this is easy. $\endgroup$ – Did Aug 1 '11 at 16:02
  • $\begingroup$ When you divide both side of the inequality by $(x+y)^r$ = $1 < x^r + y^r / (x+y)^r$ how is this equal to 1 ? $\endgroup$ – datauser Aug 1 '11 at 16:30
  • $\begingroup$ @Mayumi: I just said it made the left side $1$. You are correct that the right side is greater than $1$ $\endgroup$ – Ross Millikan Aug 1 '11 at 16:50

Now that you know why we can assume that $x+y=1$, we will look at the problem in a slightly different way, which I hope will add a little to the understanding. We want to prove that $(x+y)^r <x^r+y^r$. Since everything is positive, this is equivalent to showing that the ratio $(x^r+y^r)/(x+y)^r$ is greater than $1$.

But notice that $$\frac{x^r+y^r}{(x+y)^r}=\frac{x^r}{(x+y)^r}+\frac{y^r}{(x+y)^r}=\left(\frac{x}{x+y}\right)^r + \left(\frac{y}{x+y}\right)^r.$$

Each of the numbers $x/(x+y)$ and $y/(x+y)$ is positive and less than $1$.

If $t$ is positive and less than $1$, and $r<1$, then $t^r>t$. Thus $$\left(\frac{x}{x+y}\right)^r >\frac{x}{x+y}\qquad\text{and}\qquad \left(\frac{y}{x+y}\right)^r>\frac{y}{x+y},$$ and therefore $$\left(\frac{x}{x+y}\right)^r + \left(\frac{y}{x+y}\right)^r>\frac{x}{x+y}+\frac{y}{x+y}=1,$$ which is what we needed to prove.

The fact that if $0<t<1$ and $r<1$, then $t^r>t$ is more or less needed to complete the proof that assumes that $x+y=1$, though there are also calculus approaches. Thus the two-variable approach is really no harder than the one-variable approach, but bypasses the potentially puzzling "without loss of generality" part.

Comment about setting $x+y=1$: Let $F(x,y)=(x+y)^r$, and $G(x,y)=x^r+y^r$. Note that $F(ax,ay)=(ax+ay)^r =a^r(x+y)^r=a^rF(x,y)$. Note also that $G(ax,ay)=a^rG(x,y)$. We say that each of the functions $F(x,y)$ and $G(x,y)$ is homogeneous of degree $r$. The notion extends easily to functions of more variables. Many of the famous inequalities involve homogeneous functions. A simple example is the Arithmetic Mean Geometric Mean Inequality in three variables, $$\frac{x+y+z}{3} \ge \sqrt[3]{xyz}$$ (if $x$, $y$, and $z$ are non-negative). In this example, the functions are homogeneous of degree $1$.

The fact that in our case, $F(x,y)$ and $G(x,y)$ are homogeneous of the same degree is the real reason that we can, without loss of generality, assume that $x+y$ is anything we like. For note that if $a$ is positive then $$F(ax,ay)<G(ax,ay)\qquad\text{iff}\qquad a^rF(x,y)<a^rG(x,y)\qquad\text{iff}\qquad F(x,y)<G(x,y).$$

If we have established the inequality whenever $x+y=1$, then by multiplying each of $x$ and $y$ by $c^{1/r}$, we can obtain the inequality for any $x,y$ such that $x+y=c$.

Comment about the $r$-th power: What do we mean when we write $t^r$, say for $t>0$? This is quite a bit more complicated than it looks. We have a clear understanding of what we mean by $t^2$, or $t^5$. After a while, we develop an understanding of what we mean by something like $t^{3/4}$. For there is a unique positive number such that $s^4=t$, and then we can define $t^{3/4}$ to be the $s^3$.

After a while, we can show that familiar the laws of exponents that worked for integer powers, also work for expressions of the form $x^{p/q}$, where $p$ and $q$ are integers.

However, what do we mean, for example, by $3^{\sqrt{2}}$? Certainly it is not $3$ multiplied by itself $\sqrt{2}$ times!

There are several approaches to our quandary. One is to note that $\sqrt{2}\approx 1.41421356$ and think of $3^{1.4}$, $3^{1.41}$, $3^{1.414}$, $3^{1.4142}$, and so on. All these make sense, because the exponents can be expressed as fractions. But, intuitively, these numbers are getting closer and closer to something, and we define $3^{\sqrt{2}}$ to be that something.

However, it is more efficient to first of all define the functions $e^x$ and $\ln x$, and then define $t^r$ as $e^{r\ln t}$. Then it is not hard to show that the familiar laws of exponents work for any real exponent $r$.

A Problem-Solving Comment: The original inequality had symmetry between $x$ and $y$. Specializing to the case $x+y=1$ lets us turn the problem into a one variable problem, though with a certain loss of symmetry. In this case, the gain is probably worth it, though this post really has tried to show that the same idea can be pushed through without breaking symmetry.

The gain in going to one variable is mainly psychological. Because of the way schooling in mathematics is done, we have seen one variable problems far more often than two variable problems. But in many situations, it is useful to preserve symmetry as long as possible.

  • $\begingroup$ Very interesting! Thanks! $\endgroup$ – datauser Aug 2 '11 at 5:07

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