# Inequality 39(Christos Patilas)

Problem:

If it holds that $\displaystyle \left(\sin x+\cos x\right)^{k}\leq 8(\sin^{n}x+\cos^{n}x)$, then find the value of

$\displaystyle (n+k)_{\max}$.

1st solution:

From the Power-Mean Inequality we have that

$\displaystyle \sqrt[n]{\frac{\sin^nx+\cos^nx}{2}}\geq \sqrt{\frac{\sin^2x+\cos^2x}{2}}\Longrightarrow \sin^nx+\cos^nx\geq 2\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}$.

So, multiplying by $\displaystyle 8$ we have that

$\displaystyle 8\left(\sin^nx+\cos^nx\right)\geq 16\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}$.

Taking now in hand the left hand side, we know that

$\displaystyle \sin x+\cos x\leq \sqrt{2}\Longrightarrow \left(\sin x+\cos x\right)^{k}\leq \left(\sqrt{2}\right)^{k}$.

So, we know that $\displaystyle 16\cdot \left(\frac{\sqrt{2}}{2}\right)^{n}\geq \left(\sqrt{2}\right)^{k}$. Doing the manipulations on both sides we get that

$\displaystyle n+k\leq 8$, hence $\displaystyle (n+k)_{\max}=8$, Q.E.D.

2nd solution (An idea by Vo Quoc Ba Can):

The inequality is symmetric on $\displaystyle \sin x, \cos x$. So, we only need to find the maximum of those two constants for the values of which

$\displaystyle \sin x=\cos x$, that is $\displaystyle x=\frac{\pi}{4}$. S

o, plugging on the above inequality the value

$\displaystyle x=\frac{\pi}{4}$ we get the desired maximum result, Q.E.D.

# Inequality 38(Christos Patilas)

Problem:

If $\displaystyle a,b,c$ are angles of an acute triangle, prove that $\displaystyle \pi^{\pi}a^bb^cc^a\leq (a^2+b^2+c^2)^{\pi}$.

Solution:

The function $\displaystyle f(x)=\ln x$ is strictly concave, so from the general weighted Jensen inequality with weights the $\displaystyle a,b,c$ we have that

$\displaystyle \frac{b \ln a+c \ln b+a \ln c}{a+b+c}\leq \ln\left(\frac{ab+bc+ca}{a+b+c}\right)$, or $\displaystyle \frac{b \ln a+c \ln b+a \ln c}{\pi}\leq \ln\left(\frac{ab+bc+ca}{\pi}\right)$.

But, the last relation can be rewritten as

$\displaystyle \frac{1}{\pi}\cdot \ln(a^bb^cc^a)\leq \ln\left(\frac{ab+bc+ca}{\pi}\right)$, that is $\displaystyle \ln\left(a^bb^cc^a\right)^{1/\pi}\leq \ln \left(\frac{ab+bc+ca}{\pi}\right)$.

Removing the logarithm we get

$\displaystyle \left(a^bb^cc^a\right)^{1/\pi}\leq\frac{ab+bc+ca}{\pi}\Longrightarrow \pi^{\pi}a^bb^cc^a\leq \left(ab+bc+ca\right)^{\pi}\leq (a^2+b^2+c^2)^{\pi}$, Q.E.D.

# Inequality 37(Murray Klamkin)

Problem:

For all non-negative real numbers $\displaystyle a,b,c$ with sum $\displaystyle 2$, prove that

$\displaystyle (a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)\leq 3$.

Solution:

Assume without loss of generality that

$\displaystyle a\geq b\geq c$. Moreover, denote by $\displaystyle t,u$ the $\displaystyle \frac{a+b}{2},\frac{a-b}{2}$.

Then we get that

$\displaystyle a=t+u, b=t-u$. From the hypothesis we deduce also that $\displaystyle t\leq 1$.

Let us now transform the $\displaystyle 3$ factors of the inequality in terms of $\displaystyle t,u$.

Thus we have that:

$\displaystyle a^2+ab+b^2=(t+u)^2+(t+u)(t-u)+(t-u)^2=3t^2+u^2$ and $\displaystyle (b^2+bc+c^2)(c^2+ca+a^2)=(t^2+tc+c^2)^2-u^2(2tc-c^2-u^2+2t^2)$.

Define by $\displaystyle f\left(a,b,c\right)$ the Left Hand Side of the Inequality, that is

$\displaystyle f\left(a,b,c\right)=(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2)$.

We will now prove that $\displaystyle f(t,t,c)-f(a,b,c)\geq 0$. Denote the Left Hand Side by $\displaystyle X$. We must prove that $\displaystyle X\geq 0$.

We know that

$\displaystyle X=u^2(5t^4+4t^3c-6t^2c^2-2tc^2-c^4-u^4-u^2(t-c)^2)$.

We claim that the second factor of $\displaystyle X$ is greater than zero. Indeed, this is true as since $\displaystyle t=\max\left\{c,u\right\}$ we get that:

$\displaystyle 2t\geq t+c\Longrightarrow 4t^2\geq (t+c)^2\wedge t\geq c$.

Multiplying these $\displaystyle 2$ Inequalities we have that $\displaystyle 4t^3c\geq c^2(t+c)^2$ or $\displaystyle 4t^3c\geq t^2c^2+2tc^2+c^4$.  Adding up the $\displaystyle 5t^2c^2$ we get that

$\displaystyle 4t^3c+5t^2c^2\geq 6t^2c^2+2tc^3+c^4$.

Thus we have prove that

$\displaystyle 5t^4+4t^3c-6t^2c^2-2tc^2-c^4\geq 5t^4-5t^2c^2=5t^2(t-c)(t+c)$.

But from the last inequality we deduce that

$\displaystyle 5t^2(t-c)(t+c)\geq 5(t-c)t^3\geq 2(t-c)^4\geq u^2(t-c)^2+u^4$

due to the maximized value of $\displaystyle t$.

This completes the first scale of the proof. Now we only need to prove that if

$\displaystyle 2t+c=2$ then $\displaystyle 3t^2(t^2+tc+c^2)\leq 3$,

which is obviously true since it is of the form $\displaystyle (1-t)(3t^2-3t+1)\geq 0$, Q.E.D.

# Inequality 36(Christos Patilas)

Problem:

If $\displaystyle a,b,c$ are side lengths of a triangle, and $\displaystyle E,R$ be its area and circumradius respectively, satisfying the equality $\displaystyle a+b+c=3$, prove that

$\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(ER\right)^{\frac{1}{\sqrt{2}}}$.

Solution:

Let $\displaystyle E=\frac{abc}{4R}$.

Then the given inequality is of the form

$\displaystyle \frac{3}{a^2+b^2+c^2}\geq 2^{\sqrt{2}}\cdot \left(\frac{abc}{4}\right)^{\frac{1}{\sqrt{2}}}$.

Removing the square roots of $\displaystyle 2$ we get that

$\displaystyle \left(\frac{3}{a^2+b^2+c^2}\right)^{\sqrt{2}}\geq abc$.

Let us now introduce the $\displaystyle u,v^2,w^3$, that is

$\displaystyle3u=a+b+c,3v^2=ab+bc+ca,w^3=abc$.

Then the inequality takes the form

$\displaystyle w^3\leq u^3\left(\frac{u^2}{3u^2-2v^2}\right)^{\sqrt{2}}$.

Thus $\displaystyle w^3$ attains its maximum when $\displaystyle 2$ of the $\displaystyle \left\{a,b,c\right\}$ are equal.

So, we need only need to prove the inequality for $\displaystyle a=b$, or we need to prove that for $\displaystyle 2a+c=1$ it holds that

$\displaystyle \left(\frac{3}{2a^2+c^2}\right)^{\sqrt{2}}\geq a^2c$,

which is true, Q.E.D.

# Inequality 35(Gabriel Dospinescu)

Problem:

If $\displaystyle a_1,a_2,...,a_n$ are positive real numbers, such that $\displaystyle a_{1}\cdot a_2\cdot...\cdot a_n=1$, prove that

$\displaystyle \sum^{n}_{i=1}a_i\geq\sum^{n}_{i=1}\sqrt{\frac{a^{2}_{i}+1}{2}}$.

Solution:

From the AM-GM we deduce that

$\displaystyle a_{i}+1+\frac{2(a^{2}_{i}+1)}{a_i+1}\geq 2\sqrt{2}\sqrt{a^{2}_{i}+1}$.

So,

$\displaystyle \sum^{n}_{i=1}(a_i+1)+2\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq 2\sqrt{2}\sum^{n}_{i=1}\sqrt{a^{2}_{i}+1}$.

Divide the last relation by $\displaystyle 4$. Then we get that

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq \frac{1}{\sqrt{2}}\sum^{n}_{i=1}\sqrt{a^{2}_{i}+1}$,

or

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\geq \sum^{n}_{i=1}\sqrt{\frac{a^{2}_{i}+1}{2}}$.

So, it is enough to prove that

$\displaystyle \frac{1}{4}\sum^{n}_{i=1}(a_i+1)+\frac{1}{2}\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}\leq \sum^{n}_{i=1}a_i\Longrightarrow 3\sum^{n}_{i=1}a_i\geq 2\sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}+n$.

But it holds that,

$\displaystyle \sum^{n}_{i=1}\frac{a^{2}_{i}+1}{a_i+1}=\sum^{n}_{i=1}(a_i+1)-\sum^{n}_{i=1}\frac{2a_i}{a_i+1}$.

From this we need to prove now

$\displaystyle 3\sum^{n}_{i=1}a_i\geq 2\sum^{n}_{i=1}(a_i+1)-4\sum^{n}_{i=1}\frac{a_i}{a_i+1}+n$.

That is,

$\displaystyle \sum^{n}_{i=1}(a_i+1)+4\sum^{n}_{i=1}\frac{a_i}{a_i+1}\geq 4n$,

which is obvious from the AM-GM inequality, Q.E.D.