Help with this inequality needed for a proof in a textbook

[MidgetDwarf]

Homework Statement

Let a>0, b>0, and 0<λ<1 begiven. Show that ln(λa + (1 − λ)b) ≥ λ ln a + (1 − λ) ln b.

Relevant Equations

We were given a few exercises on a worksheet in order understand a proof in a Topology book.
The listed HW problem was 4.

Now, For problem 5, we have:
(AM-GMInequality)Let a > 0,b > 0,and 0 < λ <1 be given. Show that (a^λ)(b^(1−λ)) ≤ λa+(1−λ)b.

Note that if we prove problem 4, the proof for problem 5 follows directly. We use properties of logarithms to combine the right hand side of ln into a single logarithm. Then we raise both side of the inequality to a power of e. Which leads us to the desired inequality.

But, when I try to be prove 4 using 5, it leads to circular reasoning. Since, If I prove 5 first without using problem 4, I can do a proof by contradiction.

Assume instead that ln(λa + (1 − λ)b) < λ ln a + (1 − λ) ln b. Which leads to (a^λ)(b^(1−λ)) > λa+(1−λ)b. But by problem 5, we know that (a^λ)(b^(1−λ)) ≤ λa+(1−λ)b.

QED

I want to avoid circular reasoning and complete the problems in order. Was wondering if anyone can give me a hint and point me in the right direction on how to complete 4? I tried contraction, but I do not think this is the way to go.

Thanks.

 

[StoneTemplePython]

if you want to do 4 first, try differentiating the natural log a couple times.

Personally I'd do 5 first and show it implies 4 -- done correctly I don't see there's no contradiction.

 

[member 587159]

You are basically asked to prove that $\log$ is concave. Do you know that a twice differentiable function is concave if and only if its second derivative is non-positive (equivalently, a twice differentiable function is convex iff its second derivative is non-negative)? In that case, the concavity becomes trivial since $$\log(x)'' = \left(\frac{1}{x}\right)' = \frac{-1}{x^2} \leq 0$$

If you don't know this fact, you might want to attempt proving it (Hint: Taylor's theorem).

 

[MidgetDwarf]

Thanks to both of you for your help. I have mulled over the hint since the first response.

Thanks for reminding me of what concave is, it was something I knew 6 years ago but never used it again. What I am having trouble seeing is what function I am differentiating exactly. This is what I am thinking

Proof:

Let f(x) = ln a. So f(x)'= 0. Then f(x)''=0. (Derivative of a constant is 0); Therefore, f(x) = ln a is concave. Thus, by definition of concave, we arrive at the desired inequality. Not sure if I forced the proof. Maybe I am thinking too hard.

 

[MidgetDwarf]

whoops I should replace the a with b in my proof.

 

[Office_Shredder]

The function in question is f(x)=ln(x). That function is concave.

 

[member 587159]

Thanks to both of you for your help. I have mulled over the hint since the first response.

Thanks for reminding me of what concave is, it was something I knew 6 years ago but never used it again. What I am having trouble seeing is what function I am differentiating exactly. This is what I am thinking

Proof:

Let f(x) = ln a. So f(x)'= 0. Then f(x)''=0. (Derivative of a constant is 0); Therefore, f(x) = ln a is concave. Thus, by definition of concave, we arrive at the desired inequality.Not sure if I forced the proof. Maybe I am thinking too hard.

You should use $f(x)=\ln(x)$ and not $f(x)=\ln(a)$ (which is constant) If you can show it is concave it will follow (definition of concavity) that $\ln(tx+(1-t)y)\geq t\ln(x)+(1-t)\ln(y)$ for all $x,y>0, 0\leq t \leq 1$. This is what you are asked to prove if you take $t=\lambda, x=a, y=b$.

Refer again to post #3 for the correct calculation.

 

[StoneTemplePython]

to get the strictness of the convexity of the exponential function -- or equivalently the strict negative convexity of the natural log-- I think using the second derivative really is the way to go. If you don't care about strictness, for the record, here's the way to do it in reverse (5) $\implies$ (4).
for $p\in (0,1)$

$\exp\big(p\cdot x_1 + (1-p)x_2\big)$
$= \exp\big(p\cdot x_1\big)\exp\big((1-p)x_2\big)$
$= \exp\big( x_1\big)^p\exp\big(x_2\big)^{1-p}$
$\leq p\cdot \exp\big( x_1\big)+(1-p)\cdot\exp\big(x_2\big)$
which proves that the exponential function is convex. Now take the natural log of each side and use the substitution $y_i = e^{x_i}$, $\ln(y_i) = x_i$ which gives

$p\cdot \ln(y_1) + (1-p)\ln(y_2)$
$=p\cdot x_1 + (1-p)x_2$
$\leq \ln\Big(p\cdot \exp\big( x_1\big)+(1-p)\cdot\exp\big(x_2\big)\Big)$
$\ln\Big(p\cdot y_1+(1-p)\cdot y_2\Big)$

 

[MidgetDwarf]

Thank you. Both of you. As a result, I was able to prove Holder's and Mikowski's Inequality using my initial question and the special case that arises from it.