Let $x$ and $y$ be positive real numbers such that: $x+y^{2016}\geq 1$. Prove that $x^{2016}+y> 1-\frac{1}{100}$
Problem
Source: RMM 2016, Problem 4
Tags: algebra, inequalities, RMM, RMM 2016
sqing
28.02.2016 09:52
Problem 5？ Problem 6？ Thanks.
pi37
28.02.2016 11:30
If $y\ge 0.99$ we're done, so suppose $y\le 0.99$. Then $x\ge 1-y^{2016}\ge 1-0.99^{2016}$. Now note that
\[
\left(1+\frac{1}{99}\right)^{2016}=\left(\left(1+\frac{1}{99}\right)^{99}\right)^{2016/99}\ge 2^{2016/99}> 2^{20}
\]by Bernoulli's inequality. So
\[
x\ge 1-0.99^{2016}>1-2^{-20}>1-\frac{1}{2016\cdot 99+1}
\]We aim to show $x^{2016}>0.99$ (throwing out the $y$ term). Note that
\[
\left(1+\frac{1}{99}\right)^{1/2016}\ge 1+\frac{1}{99\cdot 2016}
\]again by Bernoulli, so
\[
x^{2016}\ge \left (\frac{2016\cdot 99}{2016\cdot 99 + 1}\right)^{2016}\ge 0.99
\]from the prior estimate, so we're done.
rkm0959
28.02.2016 11:52
Clearly it suffices to check $0<x<1$ and $0<y<0.99$. Also, we can let $x+y^{2016}=1$ as it only strengthens the condition. It suffices to show that $(1-y^{2016})^{2016}+y>0.99$. By Bernoulli's Inequality, it suffices to show that $1-2016y^{2016}+y>0.99$ Take $f(y)=y-2016y^{2016}$ - then $f'(y)=1-2016^2y^{2015}$. Therefore, $f(y)$ increases in $[0,\frac{1}{\sqrt[2015]{2016^2}}]$. Now we just need to check $f(y)>-0.01$ for $y=0$ and $y=0.99$. $y=0$ case is trivial. We now need to show that $(0.99)^{2016} < \frac{1}{2016}$. This is equivalent with $(1+\frac{1}{99})^{2016} > 2016$, but we have $$\left(1+\frac{1}{99}\right)^{2016} = \left(\left(1+\frac{1}{99}\right)^{99}\right)^{\frac{2016}{99}} \ge 2^{\frac{2016}{99}} \ge 2^{15} > 2016$$so we are done.
huricane
28.02.2016 11:52
pi37 wrote:
If $y\ge 0.99$ we're done, so suppose $y\le 0.99$. Then $x\ge 1-y^{2016}\ge 1-0.99^{2016}$. Now note that
\[
\left(1+\frac{1}{99}\right)^{2016}=\left(\left(1+\frac{1}{99}\right)^{99}\right)^{2016/99}\ge 2^{2016/99}> 2^{20}
\]by Bernoulli's inequality. So
\[
x\ge 1-0.99^{2016}>1-2^{-20}>1-\frac{1}{2016\cdot 99+1}
\]We aim to show $x^{2016}>0.99$ (throwing out the $y$ term). Note that
\[
\left(1+\frac{1}{99}\right)^{1/2016}\ge 1+\frac{1}{99\cdot 2016}
\]again by Bernoulli, so
\[
x^{2016}\ge \left (\frac{2016\cdot 99}{2016\cdot 99 + 1}\right)^{2016}\ge 0.99
\]from the prior estimate, so we're done.
I don't think your solution is correct. Bernoulli's inequality for $0\le r\le 1$ is $(1+x)^r\le 1+xr,\forall x\ge -1$,so $$\left(1+\frac{1}{99}\right)^{\frac{1}{2016}}\le 1+\frac{1}{99\cdot 2016}.$$
rkm0959
28.02.2016 12:00
We can just use the previous bound of pi37 of $x \ge 1-\frac{1}{2^{20}}$ then use $$x^{2016} \ge (1-\frac{1}{2^{20}})^{2016} \ge 1-\frac{2016}{2^{20}} > 1-\frac{3000}{1000^2} = 1-\frac{3}{1000} = 0.997 > 0.99$$
huricane
28.02.2016 14:13
For $y\ge 0.99$ the inequality is obvious.Now suppose that $y\le 0.99$.Then $$x^{2016}\ge (1-y^{2016})^{2016}\ge \left(1-\left(\frac{99}{100}\right)^{2016}\right)^{2016}$$,so it's enough to prove that $\left(1-\left(\frac{99}{100}\right)^{2016}\right)^{2016}>\frac{99}{100}$ or $$1>\left(\frac{99}{100}\right)^{2016}+\left(\frac{99}{100}\right)^{\frac{1}{2016}}.$$ By Bernoulli's Inequality we have $\left(\frac{99}{100}\right)^{\frac{1}{2016}}\le 1-\frac{1}{100\cdot 2016}$,so it's enough to prove that $\frac{1}{100\cdot 2016}>\left(\frac{99}{100}\right)^{2016}$ or $$\left(\frac{100}{99}\right)^{2015}>99\cdot 2016.$$ Now note that $2016<99^2$ and $\left(\frac{100}{99}\right)^{2015}>\left(\frac{100}{99}\right)^{2013}$,so we suffice to show that $$\left(\frac{100}{99}\right)^{671}>99.$$ But $$\left(\frac{100}{99}\right)^{671}=\left(\left(1+\frac{1}{99}\right)^{99}\right)^{\frac{671}{99}}>2^{\frac{671}{99}}=2^{\frac{61}{9}}>2^{6.7}>99.$$ This ends our proof. The problem can be solved using Bernoulli's Inequality exclusively,but I just wanted to show how weak the inequality is.So my question is: What is the best constant $c$ such that $(1-y^{2016})^{2016}+y>c$?
rkm0959
28.02.2016 14:19
Transforming $1-y^{2016}=x$, WolframAlpha gives the minimum of $x^{2016}+\sqrt[2016]{1-x}$ is around $0.997415$.
shmm
28.02.2016 18:12
What is the mean RMM??
elVerde
29.02.2016 00:03
Let $k=2016$. If $x\geq 1$ or $y\geq 1$ both inequalities are valid and trivial. Take $y\leq 1$ and it suffices to show the second inequality for $x=1-y^k\geq 0$. This becomes a one variable inequality: $x^{k}+y=(1-y^k)^k+y>1-\frac{1}{100}$.
Define $f(y)=(1-y^k)^k+y$ for $0\leq y\leq 1$. We want to prove $f(y)>1-1/100$.
We have $f'(y)=-k^2y^{k-1}(1-y^k)^{k-1}+1$. Any extrema of $f$ is attained at the boundary $y\in\{0,1\}$ or $y$ such that $f'(y)=0$. In the boundary, $f$ is $1$. The equation $f'(y)=0$ writes as $y(1-y^k)=k^{-2/(k-1)}$. Not too hard to see there are two values of $y$ which satisfy this equation: one is the maximum of $f$, the other is the minimum. For either, in particular the minimum, we find
\begin{align*}
f(y)
&=(1-y^k)^k+\frac{k^{-2/(k-1)}}{1-y^k}\\
&=(1-y^k)^k+k\frac{k^{-2/(k-1)-1}}{1-y^k}\\
&\geq (k+1)k^{\frac{k(-2-(k-1))}{(k+1)(k-1)}}\\
&=(k+1)k^{-k/(k-1)}=:M
\end{align*}
We used weighed AM-GM to get the inequality in the middle. Now we have to show that
\begin{align*}
M=\frac{k+1}{k\sqrt[k-1]{k}}>1-1/100
\end{align*}
This writes as
\begin{align*}
\frac{k+1}{k\sqrt[k-1]{k}}&>99/100\\
\Leftrightarrow\frac{1}{\sqrt[k-1]{k}}&>\frac{99k}{100(k+1)}\\
\Leftrightarrow\sqrt[2015]{2016}&<\frac{100\cdot 2017}{99\cdot 2016}=1+\frac{2017+99}{99\cdot 2016}\\
\Leftrightarrow2016&<\left(1+\frac{2017+99}{99\cdot 2016}\right)^{2015}.
\end{align*}
Because $\tfrac{2017+99}{99\cdot 2016}>\tfrac{1}{100}$ it suffices to show that
\begin{align*}
2016<\left(1+\frac{1}{100}\right)^{2015}
\end{align*}
Ugly: Taking the logarithm of both sides we want $\ln(2016)<2015\ln(1+\frac{1}{100})$. By the Taylor series expansion $\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\cdots$ for $x=\tfrac{1}{100}<1$ we find that $\ln(1+\tfrac{1}{100})\sim\tfrac{1}{100}$, hence $2015\ln(1+\frac{1}{100})\sim\tfrac{2015}{100}\sim 20$. The error in the estimate is inferior to $1$ and definitely negligible compared to the estimate $\ln(2016)<\log_2(2016)<\log_2(2048)=\underline{11}<20$.
Better: By Bernoulli's inequality $\left(1+\frac{1}{100}\right)^{2015}=\left(\left(1+\frac{1}{100}\right)^{100}\right)^{2015/100}>\left(1+1\right)^{2015/100}>2^{20}>2^{11}=2048>2016$.
sqing
29.02.2016 00:09
FabrizioFelen wrote: Let $x$ and $y$ be positive real numbers such that: $x+y^{2016}\geq 1$. Prove that $x^{2016}+y> 1-\frac{1}{100}$
Attachments:
randomusername
01.03.2016 12:25
WLOG $x,y\in[0;1]$, so let $t=2016$ and define the function
\[
f(z)=(1-z^t)^t+z.
\]Clearly, $x^{2016}+y\ge f(y)$ (and equality holds if $x=1-y^{2016}$), so it suffices to consider the minimum of $f$.
Then we have from the Chain Rule
\[
f'(z)=t(1-z^t)^{t-1}\cdot (0-tz^{t-1})+1=-t^2(z-z^{t+1})^{t-1}+1,
\]\[
f''(z)=-t^2(t-1)(z-z^{t+1})^{t-2}\cdot (1-(t+1)z^t).
\]It follows that $f''(z)\le 0$ for $z\le\zeta=:(t+1)^{-1/t}$, and hence $f$ is concave on $[0;\zeta]$, whence it admits its minimum value on $[0;\zeta]$ at $0$ or at $\zeta$. Since $f(0)=1$, it suffices to check the value of $f(z)$ in $[\zeta;1]$.
But $\zeta=2017^{-1/2016}>99/100$, because it is equivalent to $2017<(100/99)^{2016}$ which follows from $45<(100/99)^{1008}$, which follows from $7<(100/99)^{504}$, which follows from $3<(1+1/99)^{252}$, which is trivial from the Binomial theorem (or Bernoulli).
Therefore, $f(y)\ge \min\{f(0)=1,f(\zeta)\}$ for $y\in [0;\zeta]$ and $f(y)\ge y\ge \zeta>99/100$ for $y\in[\zeta;1]$ (in particular, $f(\zeta)>99/100$. This shows that $f(y)>99/100$ for all $y\in [0;1]$.
juckter
01.03.2016 20:03
Solution at the actual contest:
The inequality is trivial if $y \geq 1$. Assume $y < 1$ and take $x = c - y^{2016}$ with $c \geq 1$, then we wish to prove that $y + (c - y^{2016})^{2016} > 99/100$ and thus it suffices to check that $f(y) = y + (1 - y^{2016})^{2016} > 99/100$ for all $y$ in $(0, 1)$. As $f$ is continuous in $[0, 1]$ it suffices to check all points such that $f'(y) = 0$. We have
$$f'(y) = 1 - 2016^2(1 - y^{2016})^{2015}y^{2015}$$From which we can readily check that $y(1 - y^{2016}) = 2016^{\frac{-2}{2015}}$ is a necessary condition for $f'(y) = 0$. Now, using weighted AM-GM we get
$$\frac{1}{2017} (1 - y^{2016})^{2016} + \frac{2016}{2017} \cdot \frac{y}{2016} \geq \sqrt[2017]{\frac{y^{2016}(1 - y^{2016})^{2016}}{2016^{2016}}} = \sqrt[2017]{\frac{2016^{\frac{-2 \cdot 2016}{2015}}}{2016^{2016}}} = 2016^{-\frac{2016}{2015}}$$So it remains to check that $2017 \cdot 2016^{-\frac{2016}{2015}} > 99/100 \iff 2016^{\frac{2016}{2015}} < \frac{2017 \cdot 100}{99} \iff 2016 < \left(\frac{2017 \cdot 100}{2016 \cdot 99}\right)^{2015}$. We can easily check that $\frac{2017 \cdot 100}{2016 \cdot 99} > \frac{200}{199}$ so we just need to prove that $\left(\frac{200}{199}\right)^{2015} > 2016$. Notice now that
$$\left(\frac{200}{199}\right)^{2015} > \left(\frac{200}{199}\right)^{2013} = \left(\left(\frac{200}{199}\right)^{183}\right)^{11} > 2^{11} > 2016$$Using the inequality $\frac{200^{183}}{199^{183}} > 2$, which follows easily from the Binomial Theorem.
zhanchi_7057
19.05.2019 10:50
Maybe 0.99 is not sharp, how about 1-1/300, how do we proof it?
zhanchi_7057
19.05.2019 10:51
sqing wrote: FabrizioFelen wrote: Let $x$ and $y$ be positive real numbers such that: $x+y^{2016}\geq 1$. Prove that $x^{2016}+y> 1-\frac{1}{100}$ could you please tell how to proof "1-1/300"? thanks
naman12
28.04.2020 08:49
I'm surprised no one showed this solution: The inequality is obvious for $y\geq 0.99$, so let $y<0.99$. We get \[x\geq 1-y^{2016}\geq 1-0.99^{2016}\]so thus we have that \[x^{2016}+y\geq x^{2016}\geq (1-0.99^{2016})^{2016}\geq 1-2016\cdot 0.99^{2016}\geq 1-2016e^{-2016\cdot 0.01}=1-\dfrac {2016}{e^{20.16}}\]where we used the inequality, \[1+x\leq e^x\]which is found by using Bernoulli's Inequality: \[e^x=\lim_{n\to\infty}\left(1+\dfrac xn\right)^n\geq\lim_{n\to\infty} 1+x=1+x\]However, we get that \[e^{20.16}\geq 2.5^{20.16}\geq 2.5^{19.5}\geq (15.625)^{6.5}\geq 10^{6.5}\geq \sqrt{10}\cdot 10^6\geq 3\cdot 10^6\geq 201600\]so thus \[x^{2016}+y\geq 1-2016e^{-20.16}\geq 1-\dfrac{2016}{201600}\geq 0.99\] @above, you could probably use better bounding on $e^{20.16}$ to get the desired inequality.
GeronimoStilton
05.12.2020 07:41
For the dumbest possible approach; write \[y\ge \sqrt[2016]{1-x}\implies x^{2016}+y\ge x^{2016}+\sqrt[2016]{1-x}.\]It suffices to show \[x^{2016}+\sqrt[2016]{1-x}\ge 0.99\forall x\in (0,1).\]Substitute $\sqrt[2016]{1-x}=z$ so it is sufficient to show \[(1-z^{2016})^{2016}+z> 0.99\forall z\in (0,1).\]By Bernoulli's Inequality, we have \[(1-z^{2016})^{2016}+z\ge 1-2016z^{2016}+z.\]So we are done in the event that \[0.01> 2016z^{2016}-z,\]which occurs when $z\le \sqrt[2015]{1/2016}$. Actually, this occurs for $z\le \sqrt[2016]{1/2016}$ because it suffices to show $0.01> 1-z$ for $z\ge \sqrt[2015]{1/2016}$, which is immediate from the observation that $0.99^{100}<1/e\implies 0.99^{1100}<1/e^{11}<1/2^{11}<1/2016$. But $(1-z^{2016})^{2016}+z\ge z> 0.99$ for $z>\sqrt[2016]{1/2016}$, so we are done.
AwesomeYRY
28.07.2021 23:44
$\textbf{Claim: }$ $0.99^{2016}<\frac{1}{2016\cdot 100}$ $\textbf{Proof: }$ We will prove the dual. That $\frac{1}{0.99}^{2016} > 1.01^{2016}$ is large. We do some brute-force, and will use Bernoulli's fact of $(1+x)^n>1+nx$ throughout the entire problem. \[1.01^{2016} > (1.01^{100})^{20} > 2^{20} > 10^6 > 2016\cdot 100\]Thus, \[0.99^{2016} < \frac{1}{1.01^{2016}} < \frac{1}{2016\cdot 100}\]$\square$ Now note that if $y\geq 0.99$ we're immediately done, so we take the $y<0.99$ case, in which case \[x^{2016}+y >x^{2016} \geq (1-y^{2016})^{2016} > (1 - 0.99^{2016})^{2016}\]Then, using claim 2, since Bernoulli's work for $x\geq 1$, we have \[>(1-\frac{1}{2016\cdot 100})^{2016}> 1-2016\cdot \frac{1}{2016\cdot 100}\geq 0.99\]$\blacksquare$
CyclicISLscelesTrapezoid
24.01.2022 08:09
Sketch: Assume WLOG that $x+y^{2016}=1$, substitute $x=1-y^{2016}$ into $x^{2016}+y$. By calculus, the resulting equation is maximized when $y-y^{2017}=\sqrt[2015]{\frac{1}{2016^2}}$. If we show that $\sqrt[2015]{\frac{1}{2016^2}} \geq 0.99$, then we're done. Take the $-\frac{2015}{44}$th power of both sides so that it's sufficient to show $1.42 \leq 1.01^{45}$, but that's true by Bernoulli so we are done.
shendrew7
27.04.2024 06:05
We are finished if $y \ge .99$, so assume $y < .99$. Then \[x \ge 1-(.99)^{2016} > 1-(.6)^{31}\]\[\implies x^{2016}+y > x^{2016} > 1 - 2016 \cdot (.6)^{31} > .99. \quad \blacksquare\]
pie854
20.08.2024 09:58
Note that if $x^{2016}>0.99$ then we are trivially done. Now consider the function $$f(x)=1-x-(0.99-x^{2016})^{2016}$$defined for $x\in [0,0.99^{1/2016}]$. This function attains a minimum as it's defined over a compact set. Suppose $f'(t)=0$ then $$-1+2016^2 t^{2015}(0.99-t^{2016})^{2015}=0\Rightarrow t(0.99-t^{2016})=2016^{-2/2015} \qquad (\star)$$however $$-t^{2017}<0.99(1-t)\Rightarrow t(0.99-t^{2016})<0.99<2016^{-2/2015}$$and so $(\star)$ has no root for $t\in (0,0.99^{1/2016})$. So the minimum occurs at the endpoints. Since both $f(0)>0$ and $f(0.99^{1/2016})>0$, it follows that $f(x)>0$ for all $x\in [0,0.99^{1/2016}]$. Now we are done since $$y^{2016}\geq 1-x\geq (0.99-x^{2016})^{2016}.$$
bjump
23.12.2024 07:01
Solved with ARNOV We can assume $y<.99$ because otherwise it wouldn't be true. We can also assume $x+y^{2016}=1$ because that would minimize $x^{2016}+y$. Now by bernoulli $$(1-y^{2016})^{2016}+y \ge 1-2016y^{2016}+y$$This has derivative $-2016^2 y^{2015} +1$ which is strictly decreasing on the interval $(0,.99)$. Now it suffices to show $2016 < (\tfrac{100}{99})^{2015}$. By Bernoulli we have $$((\tfrac{100}{99})^{99})^{\tfrac{2016}{99}} \ge 2^{\tfrac{2016}{99}} > 2^{20} >>>> 2016$$Done.