# Inequality for small values of $t$

Suppose $x,(y > 0)$ are real numbers. I want to know if it is true that for small $t$, we have

$$(tx)^2 + (ty)^2 \leq 2ty$$

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$$\imp\quad x^{2} + \pars{y - {1 \over t}}^{2} \leq \pars{1 \over \verts{t}}^{2}$$ This inequality defines a region 'inside' a circle with center at $\ds{\pars{0,{1 \over t}}}$ and radius $\ds{1 \over \verts{t}}$. From that result, it's pretty obvious that we can satisfy the OP question.

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Put $2y/(x^2+y^2)$ to be $K$. Then the inequality becomes $t^2 \leq tK$ and this is true for sure when $t \leq K$ here i am assuming that t is positive else the RHS is negative and LHS is positive

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Supposing that $x$ and $y$ are fixed, rewrite to $(tx)^2+(ty)^2-2ty\le 0$, and let $f(t)$ be the left-hand side. Clearly $f(0)=0$.

If $f'(0)<0$, then sufficiently small positive $t$ will satisfy $f(t)<0$.

If you want it to hold for small $t$ of either sign, you need $f'(t)=0$ and $f''(t) < 0$.

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