# Tag Info

### Show $\sum\limits_{k=1}^n \sqrt{a_k^2+b_k^2}\ge \sqrt{(\sum\limits_{k=1}^n a_k)^2+(\sum\limits_{k=1}^n b_k )^2}$

Let $\lVert \cdot \rVert$ be Euclidean norm on $\mathbb{R}^2$, and $v_i=(a_i,b_i)$. Then, the inequality can be written as $\sum_{i=1}^n \lVert v_i \rVert \geqq \lVert \sum_{i=1}^n v_i \rVert$ ,...
• 163
Accepted

### Show that $\sqrt{\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{6b}{a}-\frac{6a}{b}+ 23}$ is a perfect square natural number.

Note that the relation $a^2 - b^2 = 3ab$ implies that \begin{align} \frac{a}{b} - \frac{b}{a} = 3. \quad (*) \end{align} On squaring the above equation, we obtain \begin{align*} \frac{a^2}{b^2} + \...
• 2,986
Accepted

• 90.7k

• 13.5k

### Show that $\sqrt{\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{6b}{a}-\frac{6a}{b}+ 23}$ is a perfect square natural number.

Set $c=\frac{a}{b}$. The given condition on $a$ and $b$ becomes $c^2=3c+1$. The expression to investigate is $$c^2+\frac{1}{c^2}+\frac{6}{c}-6c+23$$ which you can simplify using $c^2 = 3c+1$ to reduce ...
• 18.9k
1 vote

### Find the value of a for which the sum of the elements of the set $A(a) ∩ N$ is 203

The last term is $a^2+29a+201=(a+14)^2+a+5\lt (a+15)^2$ so there are $14$ terms to be consider. It follows the equation $$(a+1)+(a+2)+\cdots+(a+14)=14a+\dfrac{14\times15}{2}=203\Rightarrow 14a=98$$ ...
• 30.3k
1 vote

### Find the value of a for which the sum of the elements of the set $A(a) ∩ N$ is 203

Corrected thanks to comment by Empy2 The numbers that yield integer square roots will be of the forms $(a+1)^2,(a+2)^2,\dots,(a+k)^2$ and their square roots will be $a+1,a+2,\dots a+k$. Their sum will ...
• 7,388
1 vote
Accepted

### solution-verification | Find $x$ from the some equalities

It depends on the set of numbers from which $a$ and $n$ may be chosen. Clearly $(-1)^2=1$ and there are other solutions with certain values of $a$ on the unit circle in the complex plane; but ...
• 2,081
1 vote

### Proving $f(a)=1 \forall \; a≥ 1/8$ and its Relation to Cardano's Formula

Let $x =\left( a + \frac{\sqrt{\frac{8a - 1}{3}} \cdot \left( a + 1 \right)}{3} \right)^{\frac{1}{3}}$ and let \$y = \left( a - \frac{\sqrt{\frac{8a - 1}{3}} \cdot \left( a + 1 \right)}{3} \right)^{\...
• 3,310

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