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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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up vote 1 down vote accepted

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \pars{tx}^{2} + \pars{ty}^{2} \leq 2ty\quad\imp\quad x^{2} + \pars{y^{2} - {2 \over t}\,y + {1 \over t^{2}}} \leq {1 \over t^{2}} $$

$$ \imp\quad x^{2} + \pars{y - {1 \over t}}^{2} \leq \pars{1 \over \verts{t}}^{2} $$ 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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