Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question

3 Answers 3

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.

share|improve this answer

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

share|improve this answer

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 $f''(t) < 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.