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Here is a way using Power Means or QM-AM inequality: $$\frac{a^2+b^2}2 \le \sqrt{\frac{a^4+b^4}2} = \sqrt{\frac{a^2+b^2}2} \implies \sqrt{a^2+b^2} \le \sqrt2$$
An easier solution. Use Lagrange multipliers for $f(x,y) = x^2+y^2$ with the constraint $$g(x,y) = x^4 - x^2 + y^4 - y^2 = 0.$$ Putting $\nabla f(x,y)$ and $\nabla g(x,y)$ in rows: $$\begin{vmatrix} x & y \\ 2x^3 - x & 2y^3-y\end{vmatrix}=0 \implies xy^2-yx^3=0.$$ If $x \neq 0$ and $y \neq 0$ (the trivial case), we get $y = \pm x$. Back to $g(x,y) = ... 2 We have $$0 = x^4-x^2 + y^4-y^2 = \left(x^2-\dfrac12\right)^2 + \left(y^2-\dfrac12\right)^2 - \dfrac12 \implies \left(x^2-\dfrac12\right)^2 + \left(y^2-\dfrac12\right)^2 = \dfrac12$$ I trust you can finish off from here. 4$n+3=x^3$,$n=x^3-3$.$n^2+3n+3=(x^3-3)^2+3(x^3-3)+3=x^6-3x^3+3=(x^2-(1/x))^3+x^{-3}>(x^2-1)^3$, but also$x^6-3x^3+3<(x^2)^3$, so it's not a cube. 1 The first thing I try to do in situations like this is to sketch a graph of the function in the neighborhood where the limit is evaluated. In your case, it is not hard to see that when$x$is sufficiently small and positive,$\lfloor \sin x \rfloor = 0$, for example. So then $$f(x) = \frac{x}{x- \lfloor \sin x \rfloor}$$ in this region of$x$will simply ... 0 In the definition of "limit" choose$\delta = \frac14$. Then we will show that there is no combination of limit value$v$and "small enough"$\epsilon > 0$such that whenever$|x-0| < \epsilon$, $$\left| \frac{x}{x - \lfloor \sin x \rfloor} - v \right| < \delta$$ If we show this, then the definition of a limit cannot be met and$f(x)$has no ... 0 If$6q^2=p^2$we have an even number of threes in the prime factorization of the rhs, whereas the number of threes in the lhs is odd. 1 Suppose$\sqrt{6} = \dfrac{p}{q}$such that$gcd(p,q)=1$. This means$6q^2 = p^2$. So$3|p^2$, since 3 is prime, so it has to divide$p$. Therefore,$p^2$has a factor of$9$. Comparing the multiplicity of$3$on the LHS, we conclude$3|q^2$and thus$3|q$. Hence the contradiction. You can also argue with taking$2$as your factor instead of$3$. 0 Possible sketch for a proof, where$\Sigma=\left\{(p\land q)\leftrightarrow(r\land s),(\neg r\land q)\right\}$:$(p\land q)\leftrightarrow(r\land s)\vdash(p\land q)\to(r\land s)\quad$[$\leftrightarrow$-elimination]$\neg r\land q\vdash\neg r\quad$[$\land$-elimination]$\Sigma\vdash\neg r\lor\neg s\quad$[$\lor$-introduction 2] ... 1 For any$n \in \mathbb{N}$, we have$ n \equiv 0,1,2 \text{ mod } 3 $. Hence$ n^2 \equiv 0,1 \text{ mod } 3 $. Thus$ n^2+4 \equiv 1,2 \neq 0 \text{ mod } 3 $. So for any integer$n \in \mathbb{N}$,$ n^2+4$is not divisible by$3 \$.