Inequality 15(Kvant)

Problem:

Prove that for any real numbers \displaystyle a_1,a_2,...,a_n the following inequality holds:

\displaystyle \left(\sum^{n}_{i=1}a_{i}\right)^{2}\leq \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}.

Solution:

Observe that

\displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j}=\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt.

But \displaystyle \sum^{n}_{i,j=1}\frac{ij}{i+j-1}a_{i}a_{j} can be considered as a constant. So,

\displaystyle \sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\int^{1}_{0}t^{i+j-2}dt=\int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt.

Notice now that

\displaystyle \int^{1}_{0}\left(\sum^{n}_{i,j=1}ia_{i}\cdot ja_{j}\cdot t^{i-1+j-1}\right)dt=\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt.

So, the inequality reduces to

\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\sum^{n}_{i=1}a_{i}\right)^{2}.

Now, using Cauchy-Schwartz inequality for integrals, we get that

\displaystyle \int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)^{2}dt\geq \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}.

But, we must now observe that

\displaystyle \left(\int^{1}_{0}\left(\sum^{n}_{i=1}ia_{i}\cdot t^{i-1}\right)dt\right)^{2}=\left(\sum^{n}_{i=1}a_{i}\right)^{2},

which comes to the conclusion, Q.E.D.

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Inequality 14(Unknown Author)

Problem:

Let \displaystyle a,b,c>0 with \displaystyle a+b+c=1. Show that

\displaystyle \frac{a^2+b}{b+c}+\frac{b^2+c}{c+a}+\frac{c^2+a}{a+b}\geq 2.

Solution:

Subtract each side with \displaystyle -1. Then, from hypothesis we have that

\displaystyle \frac{a^2+b}{b+c}-a+\frac{b^2+c}{c+a}-b+\frac{c^2+a}{a+b}-c\geq 1.

Or

\displaystyle \begin{aligned}\frac{a^2+b}{b+c}-a+\frac{b^2+c}{c+a}-b+\frac{c^2+a}{a+b}-c=\frac{a(a-c)+b(b+c)}{b+c}&+\frac{b(b-a)+c(c+a)}{c+a}\\&+\frac{c(c-b)+a(a+b)}{a+b}\geq 1\end{aligned}.

From the last relation we have that

\displaystyle \sum_{cyc}\frac{a(a-c)}{b+c}+\sum_{cyc}a\geq 1\Longrightarrow \sum_{cyc}\frac{a(a-c)}{b+c}\geq 0.

Assume without loss of generality that \displaystyle 0<a\leq b\leq c.

Then we have that

\displaystyle \frac{b(b-a)}{c+a}+\frac{c(c-b)}{a+b}\geq \frac{a(c-a)}{b+c}.

Doing some manipulations on both sides we acquire that

\displaystyle b^4+b^3c-a^2b^2-a^2bc+c^4+c^3a-b^2c^2-ab^2c\geq a^2c^2+abc^2-a^4-a^3b.

Or,

\displaystyle \sum_{cyc}a^4+\sum_{cyc}a^3b\geq \sum_{cyc}a^2b^2+abc\sum_{cyc}a,

which holds.

Indeed, from the AM-GM inequality we have that

\displaystyle \sum_{cyc}a^4\geq \sum_{cyc}a^2b^2

and from Cauchy-Schwartz inequality

\displaystyle \sum_{cyc}\frac{a^2}{c}\geq \sum_{cyc}a\Longrightarrow \sum_{cyc}a^3b\geq abc\sum_{cyc}a.

Adding up these \displaystyle 2 inequalities we get the desired result, Q.E.D.

Inequality 12(Vasile Cirtoaje)

Problem:

If \displaystyle a,b,c are non-negative numbers prove that

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

Solution:

Lemma: \displaystyle 4(a^2+ab+b^2)\geq 3(a+b)^2.

Back to the inequality now, multiply both sides by \displaystyle 64. Then we have that \displaystyle 4^3\prod_{cyc}(a^2+ab+b^2)\geq 4^3(ab+bc+ca)^3.

But from the lemma we reduce the current inequality to

\displaystyle 27\prod_{cyc}(a+b)^2\geq 64(ab+bc+ca)^3.

It also holds \displaystyle (a+b+c)^2\geq 3(ab+bc+ca). Multiplying the last inequality with \displaystyle \frac{64}{3}(ab+bc+ca)^2 we get that \displaystyle \frac{64}{3}(ab+bc+ca)^2(a+b+c)^2\geq 64(ab+bc+ca)^3.

So, it suffices to prove that

\displaystyle 27\prod_{cyc}(a+b)^2\geq \frac{64}{3}(a+b+c)^2(ab+bc+ca)^2

or

\displaystyle 9(a+b)(b+c)(c+a)\geq 8(a+b+c)(ab+bc+ca),

which reduces to the obvious inequality

\displaystyle a(b-c)^2+b(c-a)^2+c(a-b)^2\geq 0.

Equality occurs for \displaystyle (a,b,c)=(1,1,1) and also for \displaystyle (a,b,c)=(1,0,0) or any cyclic permutation, Q.E.D.

Inequality 9(Christos Patilas)

Problem:

If \displaystyle a_{1},a_{2},a_{3} are the positive real roots of the equation \displaystyle 4x^3-kx^2+mx-9=0 prove that

\displaystyle k\geq 4\sqrt[3]{\sum_{cyc}\left(a_{1}\sqrt{a_{2}+a_{3}}\right)+3\prod_{cyc}(a_{1}+a_{2})}.

Solution:

Let us divide both sides by \displaystyle 4 and then cube them.

We acquire

\displaystyle \left(\frac{k}{4}\right)^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}).

But from Viete’s relations we have that

\displaystyle \frac{k}{4}=a_{1}+a_{2}+a_{3} and \displaystyle a_{1}a_{2}a_{3}=\frac{9}{4}.

So our inequality transforms into

\displaystyle (a_{1}+a_{2}+a_{3})^{3}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}),

or

\displaystyle \sum_{cyc} a^{3}_{1}+3\prod_{cyc}(a_{1}+a_{2}) \geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}+3\prod_{cyc}(a_{1}+a_{2}).

So, it suffices to prove that

\displaystyle \sum_{cyc} a^{3}_{1}\geq \sum_{cyc}a_{1}\sqrt{a_{2}+a_{3}}.

But the last inequality holds because

\displaystyle a^{3}_{1}+a^{3}_{2}+a^{3}_{3}\geq a_{1}a_{2}(a_{1}+a_{2})+a^{3}_{3}\geq 2\sqrt{a_{1}a_{2}a_{3}\cdot a^{2}_{3}(a_{1}+a_{2})}=3a_{3}\sqrt{a_{1}+a_{2}}.

Adding up the \displaystyle 3 cyclic relations we come to the desired inequality, Q.E.D.