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  1. Prove that $ \sqrt{\frac{2x^2-2x+1}{2}}\geq\frac{1}{x+\frac{1}{x}} $ for $ 0 < x < 1. $

This one seems reminiscent of the quadratic mean on the left, maybe $\sqrt{\frac{(x-1)^2+x^2}{2}}$, but I can't find a way to compare it to the LHS, which loos a bit like the harmonic mean.

  1. Prove $ \frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ for positive reals $x,y,z$

This one seems a LOT like Cauchy-Schwarz, but I can't quite figure out which lists to use.

I need to solve these problems, but I am pretty stuck. Hints or even solutions would be greatly appreciated. Thanks for the help in advance!

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That's two separate problems.

1) Lets try $QM \ge AM$ $$LHS=\sqrt{\frac{(1-x)^2+x^2}2} \ge \frac{\overline{1-x}+x}2=\frac12 > \frac1{x+\frac1{x}} = RHS$$ as $x+\frac1x > 2$ for $x \in (0, 1)$. Note that this inequality remains strict.


2) From $x^2+y^2\ge \frac12(x+y)^2 \implies$: $$\sum_{cyc} \frac{x+y}{x^2+y^2} \le 2\sum_{cyc} \frac1{x+y} \le 2\sum_{cyc}\frac14\left(\frac1x+\frac1y\right)= \sum_{cyc} \frac1x$$ where we have used Cauchy-Schwarz also. Equality is when $x=y=z$.

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For the first question, you can try as follows. Since \begin{align*} \sqrt{\frac{2x^2-2x+1}{2}}-\frac{x}{x^2+1}=\frac{1}{2}\cdot \frac{\sqrt{4x^2-4x+2}\cdot (x^2+1)-2x}{x^2+1}=:f(x), \end{align*} if we set \begin{gather*} a(x):=\sqrt{4x^2-4x+2}=\sqrt{4\left(x-\frac{1}{2}\right)^2+1}, \end{gather*} the numerator of $f(x)$ is \begin{align*} a(x) \cdot x^2-2x+a(x)&=a(x)\cdot \left(x-\frac{1}{a(x)}\right)^2+a(x)-\frac{1}{a(x)}\\ &=a(x)\cdot \left(x-\frac{1}{a(x)}\right)^2+\frac{4\left(x-\frac{1}{2}\right)^2}{a(x)}\\ &> 0, \end{align*} from which we can conclude that $f(x)\geq 0, $ that is, \begin{gather*}\tag{$\star$} \sqrt{\frac{2x^2-2x+1}{2}}> \frac{x}{x^2+1}, \qquad \forall x\in\mathbb{R}. \end{gather*} From $(\star)$ we have \begin{gather*} \sqrt{\frac{2x^2-2x+1}{2}}>\frac{1}{x+\frac{1}{x}}, \qquad \forall x\in\mathbb{R}\backslash\{0\}. \end{gather*}

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