Inequality in real analysis $x^a y^{1-a} \le ax+ (1-a)y$

Question

How to prove that for $a\in (0,1)$ and for all $x,y >0$ we get the inequality $$x^a y^{1-a} \le ax+ (1-a)y.$$ It seems not so difficult, yet I'm stuck. I tried to obtain it from Taylor's series $$x^a y^{1-a} =y\left(1+\frac{x-y}{y}\right)^a \leq y\left(1+ a\cdot \frac {x-y}{y}\right)=ax+ (1-a)y$$ but it only works when $|\frac{x-y}{y}|<1$. Should I consider many cases, eg. $x<0$, $y<0 ,\ldots$?

Answer 1

Note that $t\to\ln(t)$ is an increasing and concave function in $(0,+\infty)$. Therefore $$\ln(x^a y^{1-a})=a\ln(x)+(1-a)\ln(y) \le \ln(ax+ (1-a)y).$$

Answer 2

Use the convexity of exponential function, observe that every positive real number has a logarithm, i.e $\exists \ p, q \in \mathbb{R}$ such that $x=e^p$ and $y=e^q$.

Therefore we are required to prove $e^{ap + (1-a)q} \le ae^{p}+(1-a)e^q$, which is the definition of convexity. Now the convexity of exponential can be proved by using the fact that second derivative is positive.

Answer 3

It's just AM-GM: $$ax+(1-a)y\geq x^ay^{1-a}$$

AM-GM it's the following thing.

Let $x_i>0$, $\alpha_i>0$ such that $\alpha_1+\alpha_2+...+\alpha_n=1$. Prove that: $$\alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n\geq x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$$

Answer 4

If you really want to use real analysis, you can move all the terms to one side like this: $$ax + (1-a)y - x^a y^{1-a} \geq 0$$ and look for the extremum of the left-hand side. You'll get $\partial / \partial x = a(1 - x^{a-1})$ and $\partial/\partial y= (1-a)(1 - y^{-a})$. We see that there is an extremum in $x = y = 1$. In this point, the left-hand side is truly equal to 0. So far OK.

The last thing is to compute the Hessian matrix in the point $x = y = 1$ and check that it's positive definite (so the 0 is a minimum). You'll get $$\begin{pmatrix} a(1-a)x^{a-2} & 0 \\ 0 & a(1-a)y^{-a-1}\end{pmatrix}.$$ For $x = y = 1$: $$\begin{pmatrix} a(1-a) & 0 \\ 0 & a(1-a)\end{pmatrix}.$$ That's obviously positive-definite (because $0 <a < 1$).

Of course this is quite a dumb way. The accepted answer is much nicer. This has just an advantage of being quite universal since it doesn't rely on knowledge of other inequalities nor on clever ideas :--).

Answer 5

Here is one approach based on mean value theorem and the credit for this proof goes to G. H. Hardy.

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Note that there is equality if $x=y$ and we can assume $x<y$ otherwise just put $b=1-a$ and let $b$ play the role of $a$. Let $f(x) = x^{1-a}$ so that $f'(x) =(1-a) x^{-a} $. Now by mean value theorem we have $$f(y) - f(x)=(y-x) f'(c) $$ for some $c$ with $x<c<y$. Therefore $$y^{1-a}-x^{1-a}=(y-x)(1-a)c^{-a}<(y-x) (1-a)x^{-a}$$ Multiplying by $x^{a} $ we get $$x^{a}y^{1-a}-x<(1-a)y-(1-a)x$$ or $$ x^{a}y^{1-a}<ax+(1-a)y$$ as desired.