#5 Jensen's Inequality Intro

The function \(x^2\) is convex. Jensen gives \(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\). Multiplying by \(3\) gives the result.

By Jensen for the concave function \(\sqrt{x}\), \[\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\le\sqrt{\frac{x+y+z}{3}}=\sqrt{9}=3.\] Hence the sum of roots is at most \(9\).

By Jensen, \[\frac13\left(\frac1a+\frac1b+\frac1c\right)\ge\frac{1}{(a+b+c)/3}=\frac{3}{a+b+c}.\] Multiply by \(3\).

Since \(\ln x\) is concave, \[\frac{\ln a+\ln b+\ln c}{3}\le\ln\frac{a+b+c}{3}=\ln2.\] Hence \(\ln(abc)\le3\ln2=\ln8\), so \(abc\le8\).

We have \(x^2-2x+1=(x-1)^2\ge0\). This says exactly that the graph of the convex function \(x^2\) lies above its tangent at \(1\).

Since \(x>0\), the inequality is equivalent to \(1\ge2x-x^2\), or \((x-1)^2\ge0\). This is the tangent to \(1/x\) at \(1\).

Since \(f(t)=t^2\) is convex, \[p f(x)+(1-p)f(y)\ge f(px+(1-p)y).\] This is exactly the required inequality.

By the concavity of \(\sqrt{x}\), \[\sqrt{\frac{x+2y}{3}}\ge\frac13\sqrt{x}+\frac23\sqrt{y}.\] Multiply by \(3\).

By Jensen, \[\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\le\sqrt{\frac{48}{3}}=4.\] Hence the sum is at most \(12\). Equality occurs at \(x=y=z=16\), so the maximum is \(12\).

Since \(\ln x\) is concave, \[\frac{\ln a+\ln b+\ln c}{3}\le\ln\frac{a+b+c}{3}=\ln4.\] Therefore \(\ln(abc)\le\ln64\), so \(abc\le64\). Equality holds at \(a=b=c=4\).

By Jensen, \[\frac13\sum\frac{1}{1+a}\ge\frac{1}{1+(a+b+c)/3}=\frac{1}{3}.\] Multiplying by \(3\), we get the result.

The function \(f(x)=\frac{x}{1+x}\) is concave for \(x\ge0\). Jensen gives \[\frac13\sum\frac{a}{1+a}\le \frac{1}{1+1}=\frac12.\] Hence the sum is at most \(\frac32\).

By Jensen, \[\frac{a^4+b^4+c^4}{3}\ge\left(\frac{a+b+c}{3}\right)^4.\] Multiplying by \(3\) gives \(\sum a^4\ge\frac{(a+b+c)^4}{27}\).

By Jensen, \[\frac{e^x+e^y+e^z}{3}\ge e^{(x+y+z)/3}=e^0=1.\] Therefore the sum is at least \(3\).

The function \(\ln x\) is concave, so its graph lies below every tangent. At \(1\), the tangent has equation \(y=x-1\). Hence \(\ln x\le x-1\).

Since \(\sqrt{x}\) is concave, its graph lies below the tangent at \(1\). This tangent is \(y=1+\frac12(x-1)=\frac{x+1}{2}\). Hence \(\sqrt{x}\le\frac{x+1}{2}\).

The function \(\ln(1+t)\) is concave for \(t\ge0\). Thus \[\frac{\ln(1+x)+\ln(1+y)+\ln(1+z)}{3}\le\ln\left(1+\frac{x+y+z}{3}\right)=\ln2.\] Hence the logarithm of the product is at most \(\ln8\), so the product is at most \(8\).

The function \(f(t)=\frac1{2+t}\) is convex for \(t\ge0\). Jensen gives \[\frac13\sum\frac1{2+x}\ge \frac1{2+(x+y+z)/3}=\frac13.\] Multiply by \(3\).

By Jensen, \[\frac{x^4+y^4+z^4}{3}\ge\left(\frac{x+y+z}{3}\right)^4=1.\] Thus the sum of fourth powers is at least \(3\). Equality holds at \(x=y=z=1\).

The function \(f(t)=t\ln t\) is convex for \(t>0\). Jensen gives \[\frac{x\ln x+y\ln y+z\ln z}{3}\ge f\left(\frac{x+y+z}{3}\right)=f(1)=0.\] Therefore \(\ln(x^x y^y z^z)\ge0\), so \(x^x y^y z^z\ge1\).

The function \(e^x\) is convex on the whole real line. Therefore Jensen says that the average of the function values is at least the function value at the average argument: \[\frac1n\sum e^{x_i}\ge e^{(\sum x_i)/n}.\]

The function \(\ln(1+x)\) is concave for \(x\ge0\). Hence \[\frac1n\sum_{i=1}^n\ln(1+x_i)\le\ln\left(1+\frac{S}{n}\right).\] Multiplying by \(n\), we get \(\ln\prod(1+x_i)\le n\ln(1+S/n)\). Exponentiating gives the result.

Since \(f(t)=t\ln t\) is convex for \(t>0\), Jensen gives \[\frac1n\sum_{i=1}^n x_i\ln x_i\ge f\left(\frac{x_1+\cdots+x_n}{n}\right)=f(1)=0.\] Hence \(\ln\prod x_i^{x_i}=\sum x_i\ln x_i\ge0\). Therefore \(\prod x_i^{x_i}\ge1\).

By Jensen, \[\frac1n\sum_{i=1}^n\frac1{x_i^2}\ge \frac{1}{\left((x_1+\cdots+x_n)/n\right)^2}=1.\] Multiplying by \(n\), we get the result.