Inequalities and…inequalities

Inequality 53 (Vo Quoc Ba Can) October 16, 2011

Filed under: Uncategorized — Basdekis George @ 5:05 pm

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) October 16, 2011

Filed under: Uncategorized — Basdekis George @ 4:52 pm

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.

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Inequality 51 (George Basdekis) October 15, 2011

Filed under: Uncategorized — Basdekis George @ 6:40 pm

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.

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Inequality 50 (Vasile Cirtoaje) October 15, 2011

Filed under: Uncategorized — Basdekis George @ 6:28 pm

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.

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Inequality 49 (Unknown Author) October 3, 2011

Filed under: Uncategorized — Basdekis George @ 6:10 pm

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

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

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

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.

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Inequality 48 (George Basdekis) October 2, 2011

Filed under: Uncategorized — Basdekis George @ 4:52 pm

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.

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Inequality 47(Christos Patilas) May 13, 2010

Filed under: Uncategorized — Basdekis George @ 2:10 pm

Problem:

If $\displaystyle x_i$ for $\displaystyle i=1,2,...,n$ are positive real numbers then prove that

$\displaystyle \sum^{n}_{i=1}\left(5\sqrt[5]{x^{3}_{i}}-3\sqrt[3]{\left(\frac{3x_i+2}{5}\right)^5}\right)\leq 2n$

Solution(An Idea by Vo Quoc Ba Can):

We only need to prove that

$\displaystyle 5\sqrt[5]{a^3}-3\sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\leq 2$ for all $\displaystyle a>0$ .

So, using the AM-GM Inequality we have that

$\displaystyle a+a+a+1+1\geq 5\sqrt[5]{a^3}$ .

It follows that

$\displaystyle \sqrt[3]{\left(\frac{3a+2}{5}\right)^5}\geq \sqrt[3]{\left(\sqrt[5]{a^3}\right)^5}=a$ .

Therefore it suffices to prove that

$\displaystyle 5\sqrt[5]{a^3}-2\leq 3a$ ,

which is obviously true from the AM-GM Inequality, Q.E.D.

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Inequality 46(George Basdekis) April 11, 2010

Filed under: Uncategorized — Basdekis George @ 1:38 pm

Problem:

Let $\displaystyle a,b,c$ be positive real numbers. Prove that $\displaystyle \left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)-8abc\geq a(b-c)^2+b(c-a)^2+c(a-b)^2$ .

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Inequality 45(Vo Quoc Ba Can) March 24, 2010

Filed under: Uncategorized — Basdekis George @ 3:26 pm

Problem:

Let $\displaystyle a,b,c$ be positive real numbers such that $\displaystyle ab+bc+ca=1$ . Prove that $\displaystyle \frac{1}{\sqrt{1+(2a-b)^2}}+\frac{1}{\sqrt{1+(2b-c)^2}}+\frac{1}{\sqrt{1+(2c-a)^2}}\leq\frac{3\sqrt{3}}{2}$ .