Inequality 39(Christos Patilas)


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)


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}.


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)


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.


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)


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}}}.


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


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)


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}}.


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}.


\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},


\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.