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Proof of “Fermat’s last theorem”

Let x,y,z,n\in \mathbb{N}^* and n \geq z. Prove that the equation

    \[x^n+y^n=z^n\]

has no solution.

Solution

Without loss of generality , assume that x<y. If x^n+y^n=z^n held , then it would be z^n > y^n thus z^n \geq (y+1)^n. It follows from Bernoulli’s inequality that,

    \begin{align*} z^n &=x^n +y^n\\ &< 2y^n \\ &\leq \left(1 + \frac {n}{z} \right) y^n \\ &\leq \left(1 + \frac {1}{z} \right)^ny^n\\ &< \left(1 + \frac {1}{y} \right)^ny^n \\ &= (y+1)^n \\ &\leq z^n \end{align*}

which is an obscurity. The result follows.

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Double binomial sum

Evaluate the double sum

    \[\mathcal{S} = \sum_{i=0}^m \sum_{j=0}^n \frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}\]

Solution

We sum diagonally , hence:

    \begin{align*} \sum_{i=0}^{m}\sum_{j=0}^{n}\frac{\binom{m}{i}\binom{n}{j}}{\binom{m+n}{i+j}}&\overset{i+j=N}{=\! =\! =\! =\!}\sum_{N=0}^{m+n}\sum_{i=0}^m\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &\stackrel{(1)}{=}\sum_{N=0}^{m+n}\sum_{i=0}^{N}\frac{\binom{m}{i}\binom{n}{N-i}}{\binom{m+n}{N}}\\ &=\sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}\\ &\stackrel{(2)}{=} \sum_{N=0}^{m+n}\binom{m+n}{N}^{-1}\binom{m+n}{N}\\ &=m+n+1 \end{align*}

(1): For i=m+1,\ldots,N it holds \binom{m}{i}=0.

(2): \displaystyle \sum_{i=0}^{N}\binom{m}{i}\binom{n}{N-i}=[x^N](1+x)^{m+n}=\binom{m+n}{N}.

Conjecture: Does the following equality

    \[\sum_{a_1=0}^{b_1}\sum_{a_2=0}^{b_2}\cdots\sum_{a_n=0}^{b_n}\frac{\binom{b_1}{a_1}\binom{b_2}{a_2}\cdots\binom{b_n}{a_n}}{\binom{b_1+b_2+\cdots+b_n}{a_1+a_2+\cdots+a_n}}=b_1+b_2+\cdots+b_n+1\]

hold?

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Kinda Pythagorean Theorem

Let {\rm AB \Gamma} be a triangle such that {\rm AB} = 10 , {\rm B \Gamma} = 22 and \hat{\rm B} = 2 \hat{\Gamma}. Find the area of the triangle.

Solution

Since \hat{\rm A} + \hat{\rm B} +\hat{\Gamma} = \pi \Rightarrow \hat{\rm A} + 3 \hat{\Gamma} = \pi it follows from the law of sines that

    \begin{align*} \frac{\alpha}{\sin \hat{\rm A}} = \frac{\beta}{\sin \hat{\rm B}} = \frac{\gamma}{\sin \Gamma} &\Rightarrow \frac{{\rm B \Gamma}}{\sin \hat{\rm A}} = \frac{\mathrm{AB}}{\sin \hat{\Gamma}}\\ &\Rightarrow \frac{22}{\sin \left ( \pi - 3 \hat{\Gamma} \right )} = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow \frac{22}{\sin 3\hat{\Gamma}} = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow \frac{22}{\sin \hat{\Gamma} \left ( 2 \cos 2 \hat{\Gamma} +1 \right ) } = \frac{10}{\sin \hat{\Gamma}} \\ &\Rightarrow 22 = 10 \left ( 2 \cos 2 \hat{\Gamma} + 1 \right ) \\ &\Rightarrow 12 = 20 \cos 2 \hat{\Gamma} \\ &\Rightarrow \frac{3}{5} = \cos \hat{\rm B} \end{align*}

Hence \displaystyle{\sin \hat{\rm B} = \sqrt{1- \cos^2 \hat{\rm B}} = \sqrt{1-\frac{9}{25}} = \frac{4}{5}}. Thus,

    \begin{align*} \mathcal{A} &= \frac{1}{2}\cdot 10 \cdot 22 \cdot \sin \hat{\rm B} \\ &=10 \cdot 11 \cdot \frac{4}{5} \\ &= 2 \cdot 4 \cdot 11\\ &= 88 \quad \text{square meters} \end{align*}

To completely justify the title of the post we give another solution based on the following proposition:

Proposition: Let ABC be a given triangle such that \hat{A} = 2 \hat{C}. Prove that

    \[a^2 = c^2 + bc\]

Proof: We are working on the following figure.

Rendered by QuickLaTeX.com

Let AD be the bisector of B\hat{A}C. Then:

    \begin{align*} \left\{ \begin{gathered} \vartriangle ADB \sim \vartriangle ABC \\ AD = DC \\ \end{gathered} \right. &\Rightarrow \left\{ \begin{gathered} \frac{{AD}} {{AC}} = \frac{{AB}} {{BC}} \\ AD = DC \\ \end{gathered} \right. \\ &\Rightarrow \frac{{DC}} {{AC}} = \frac{{AB}} {{BC}} \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{{DC = \frac{{a \cdot b}} {{b + c}},AC = B,AB = c,BC = a}}{=\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\!=\!\Rightarrow } \frac{{\frac{{a \cdot b}} {{b + c}}}} {b} = \frac{c} {a} \\ &\Rightarrow \frac{a} {{b + c}} = \frac{c} {a} \\ &\Rightarrow a^2 = c^2 + bc \end{align*}

and the result follows.

Now, the area of the initial triangle is given by Heron’s formula.

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An inequality

Let x, y \in \mathbb{R}. Prove that

    \[-\frac{1}{2}\leq \frac{\left ( x+y \right )\left ( 1-xy \right )}{\left ( 1+x^{2} \right )\left ( 1+y^{2} \right )}\leq \frac{1}{2}\]

Solution

Lemma: If a \neq 0 or b \neq 0 then it holds

    \[\frac{ab}{a^2+b^2} \leq \frac{1}{2}\]

Proof: Straightforward!

Then,

    \begin{align*} \left| {\frac{{(x + y)(1 - xy)}}{{(1 + {x^2})(1 + {y^2})}}} \right| &= \left| {\frac{{(x + y)(1 - xy)}}{{{{(x + y)}^2} + {{(1 - xy)}^2}}}} \right| \\ &\leq \frac{1}{2} \end{align*}

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Trigonometric inequality on an acute triangle

Prove that in any acute triangle ABC the following inequality holds:

    \[\frac{\sin A \sin B}{\cos C} + \frac{\sin B \sin C}{\cos A} + \frac{\sin C \sin A}{\cos B} \geq \frac{9}{2}\]

Solution

Since A+B+C = \pi it holds that

(1)   \begin{equation*} \cos C = -\cos (A +B ) = -\cos A \cos B + \sin A \sin B \end{equation*}

and thus

(2)   \begin{equation*} \frac{\sin A \sin B}{\cos C} = 1 + \frac{\cos A \cos B}{\cos C} = 1 + \frac{\tan C}{\tan A + \tan B} \end{equation*}

Using Nesbitt’s inequality we see that

    \begin{align*} \sum \frac{\sin A \sin B}{\cos C} &= \sum \left ( 1 + \frac{\tan C}{\tan A + \tan B} \right ) \\ &=3 + \sum \frac{\tan C}{\tan A + \tan B} \\ &\geq 3 + \frac{3}{2} \\ &= \frac{9}{2} \end{align*}

Equality holds if-f A=B=C=\frac{\pi}{3}.

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