Inequality 53 (Vo Quoc Ba Can)

Problem:

Let \displaystyle a,b,c>0 such that \displaystyle a+b+c=3 and \displaystyle a^2+b^2+c^2=4. Find the minimum and the maximum value of the expression

\displaystyle Q=\frac{a}{b}.

Solution:

From the Cauchy – Schwarz Inequality we have that

\displaystyle (a^2+b^2+c^2)\cdot\left[\frac{(a+b)^2}{a^2+b^2}+1\right]\geq (a+b+c)^2,

or due to the hypothesis

\displaystyle \frac{(a+b)^2}{a^2+b^2}\geq \frac{5}{4}.

From this last Inequality, we acquire

\displaystyle a^2+b^2\leq 8ab,

Let us now divide with \displaystyle ab. Then we get that

\displaystyle \frac{a}{b}+\frac{b}{a}\leq 8.

Solving this Inequality, gives us

\displaystyle 4-\sqrt{15}\leq \frac{a}{b}\leq 4+\sqrt{15}.

Thus, \displaystyle Q_{\min}=4-\sqrt{15} and \displaystyle Q_{\max}=4+\sqrt{15}, Q.E.D.

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Inequality 52 (Unknown Author)

Problem:

Let \displaystyle x_1,x_2,...,x_n be positive real numbers with sum equal to \displaystyle 1. Find the minimum value of the expression

\displaystyle A=\max\left\{\frac{x_1}{1+x_1},\frac{x_2}{1+x_1+x_2},...,\frac{x_n}{1+x_1+x_2+...+x_n}\right\}.

Solution:

From the definition of \displaystyle A, we have that

\displaystyle A\geq \frac{x_1}{1+x_1}, A\geq \frac{x_2}{1+x_1+x_2},...,A\geq \frac{x_n}{1+x_1+x_2+...+x_n}.

Solving the 1st Inequality we get that

\displaystyle x_1\leq \frac{A}{1-A}.

Solving now the 2nd Inequality we have that

\displaystyle A\geq \frac{x_2}{1+x_1+x_2}\geq \frac{x_2}{1+\frac{A}{1-A}+x_2}=\frac{x_2}{\frac{1}{1-A}+x_2}.

The last relation reduces to the

\displaystyle x_2\leq \frac{A}{(1-A)^2}.

So, we see now that \displaystyle x_k\leq \frac{A}{(1-A)^k} for \displaystyle k=1,2,...,n.

Summing up these \displaystyle n Inequalities we acquire

\displaystyle x_1+x_2+...+x_n\leq \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}

or, due to the hypothesis we now have that

\displaystyle \frac{A}{1-A}+\frac{A}{(1-A)^2}+...+\frac{A}{(1-A)^n}\geq 1.

If we solve this last Inequality it gives us the result \displaystyle A\geq 1-\frac{1}{\sqrt[n]{2}}, which is also and the value we are searching, Q.E.D.

Inequality 51 (George Basdekis)

Problem:

If \displaystyle a,b,c>0 then prove that

\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\leq \frac{3}{\sqrt{2}}\cdot\sqrt{\frac{a^2+b^2+c^2}{a+b+c}}.

Solution:

From the Cauchy – Schwarz Inequality we have that

\displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \left[\sum a(b+c)\right]\cdot\sum\frac{a}{(a+b)(b+c)}.

Expanding the RHS, we get that

\displaystyle \begin{aligned}\left(\sum\frac{a}{\sqrt{a+b}}\right)^2& \leq 2(ab+bc+ca)\cdot \frac{a^2+b^2+c^2+ab+bc+ca}{(a+b)(b+c)(c+a)}\\&\leq 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\end{aligned},

from the obvious \displaystyle ab+bc+ca\leq a^2+b^2+c^2.

Moreover, notice that the following beautiful Inequality holds, that is

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

Thus, we have that

\displaystyle 2(ab+bc+ca)\cdot \frac{2(a^2+b^2+c^2)}{(a+b)(b+c)(c+a)}\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}.

We have proved our Inequality, because \displaystyle \left(\sum\frac{a}{\sqrt{a+b}}\right)^2\leq \frac{9(a^2+b^2+c^2)}{2(a+b+c)}, or

\displaystyle \sum\frac{a}{\sqrt{a+b}}\leq 3\sqrt{\frac{a^2+b^2+c^2}{2(a+b+c)}}, Q.E.D.

Inequality 50 (Vasile Cirtoaje)

Problem:

For all \displaystyle a,b,c>0 prove that

\displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}+\frac{3b^2-2bc-c^2}{b^2+c^2}+\frac{3c^2-2ca-a^2}{c^2+a^2}\geq 0.

Solution:

It holds that \displaystyle \frac{3a^2-2ab-b^2}{a^2+b^2}=\frac{2(a^2-b^2)+(a-b)^2}{a^2+b^2}. Thus, we only need to prove that

\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq -2\sum\frac{a^2-b^2}{a^2+b^2}.

Observe that

\displaystyle \sum\frac{a^2-b^2}{a^2+b^2}=-\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}.

From the above result we have to prove now that

\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}.

Using the AM – GM Inequality we get that

\displaystyle \sum\frac{(a-b)^2}{a^2+b^2}\geq 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}.

So, it is enough to prove that

\displaystyle 3\sqrt[3]{\frac{(a-b)^2(b-c)^2(c-a)^2}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}}\geq 2\frac{(a^2-b^2)(b^2-c^2)(c^2-a^2)}{(a^2+b^2)(b^2+c^2)(c^2+a^2)}.

The last Inequality can be reduced to the \displaystyle 27(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2\geq 8(a-b)(b-c)(c-a)(a+b)^3(b+c)^3(c+a)^3.

So, it is enough to check the Inequality

\displaystyle 3(a^2+b^2)^2\geq 2|a-b|(a+b)^3.

This one can be prove by the following way. Set \displaystyle a^2+b^2=x and \displaystyle a^2-b^2=y, with \displaystyle x\geq y and remake the Inequality to the form

\displaystyle 3\frac{(a^2+b^2)^2}{(a+b)^2}\geq 2|a^2-b^2|.

Then, the Inequality substitutes to

\displaystyle 3\frac{x^2}{x+\sqrt{x^2-y^2}}\geq 2y.

But this one holds because we have that

\displaystyle 2xy++2y\sqrt{x^2-y^2}\leq 2xy+(x^2-y^2)+y^2=x^2+2xy\leq 3x^2,

due to the AM – GM Inequality and the hypothesis \displaystyle x \geq y, Q.E.D.

P.S The following nice Inequality also holds:

\displaystyle \sum \frac{3a^2-2ab-b^2}{3a^2+2ab+3b^2}\geq 0.

This Inequality belongs to Thomas Mildorf and the proof for this Inequality is the same as the above.

Inequality 49 (Unknown Author)

Problem:

Let \displaystyle a,b,c be positive real numbers such that \displaystyle a+b+c=3, and also let \displaystyle n\geq 12 be a natural number. Prove that

\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{3}{\sqrt[n-4]{n^n}}.

Solution:

Observe that \displaystyle \frac{1}{3a+bc+\frac{n-4}{a}}=\frac{a}{3a^2+bc+(n-4)}=\frac{a}{a(a+b)(a+c)+(n-4)}. Thus, from the AM – GM Inequality we get that

\displaystyle 4\cdot\frac{a(a+b)(a+c)}{4}+(n-4)\geq n\sqrt[n]{\frac{a^4(a+b)^4(a+c)^4}{4^4}}.

From the above Inequality, we are now capable of building our problem. Specifically, we have that

\displaystyle \frac{a}{a(a+b)(a+c)+(n-4)}\leq \frac{1}{n}\sqrt[n]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}

or

\displaystyle \left(\frac{a}{a(a+b)(a+c)+(n-4)}\right)^{\frac{n}{n-4}} \leq \frac{1}{\sqrt[n-4]{n^n}}\sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}.

We must see now that

\displaystyle \sqrt[n-4]{\frac{4^4a^{n-4}}{(a+b)^4(a+c)^4}}=\sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}.

From the above result we can apply once again the AM – GM Inequality and acquire

\displaystyle \sqrt[n-4]{\left(\frac{2a}{a+b}\right)^4\cdot \left(\frac{2a}{a+c}\right)^4\cdot a^{n-12}}\leq \frac{4\frac{2a}{a+b}+4\frac{2a}{a+c}+(n-12)a}{n-4}.

Summing up the \displaystyle 3 Inequalities together we get that

\displaystyle \sum_{cyc}\left(\frac{1}{3a+bc+\frac{n-4}{a}}\right)^{\frac{n}{n-4}}\leq \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}.

Moreover, we have that

\displaystyle \frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{4\sum_{cyc}\left(\frac{2a}{a+b}+\frac{2a}{a+c}\right)+3(n-12)}{n-4}=\frac{1}{\sqrt[n-4]{n^n}}\cdot\frac{3\cdot 4\cdot 2+3(n-12)}{n-4},

which reduces to the \displaystyle \frac{3}{\sqrt[n-4]{n^n}}, Q.E.D.

Inequality 48 (George Basdekis)

Problem:

If \displaystyle a,b,c,d,e,f\geq 0 such that \displaystyle a+b+c+d+e+f=6 and \displaystyle a^2+b^2+c^2+d^2+e^2+f^2=\frac{36}{5}, the prove that

\displaystyle a^3+b^3+c^3+d^3+e^3+f^3\leq \frac{264}{25}.

Solution:

Using the Cauchy – Schwarz Inequality we get that

\displaystyle \frac{36}{5}=a^2+(b^2+...+f^2)\geq a^2+\frac{(b+...+f)^2}{5}=a^2+\frac{(6-a)^2}{5}=\frac{36}{5}+\frac{6a^2-12a}{5}.

From the above Inequality we can see that \displaystyle a\leq 2. Similarly we acquire \displaystyle b,...,f\leq 2.

Observe now that

\displaystyle \sum_{cyc} a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a).

Using once again the Cauchy – Schwarz Inequality we get that

\displaystyle \sum_{cyc} a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}.

Using the hypothesis, we can calculate the last fraction. Thus we have that

\displaystyle \sum_{cyc}a^2(2-a)\geq \frac{\left[\sum_{cyc}a(2-a)\right]^2}{\sum_{cyc}(2-a)}=\frac{96}{25}.

Returning to previous Inequality, we now can see that

\displaystyle \sum_{cyc}a^3=2\sum_{cyc} a^2-\sum_{cyc}a^2(2-a)\leq 2\cdot \frac{36}{5}-\frac{96}{25}=\frac{264}{25}, Q.E.D.