show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$

Question

let $x,y>0$ and $n$ be positive integer.if $$x^{2n+1}+y^{2n+1}\ge 2$$ show that $$x^{n+1}+y^{n+1}\ge x^n+y^n$$

maybe use Holder inequality: for example $$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$ so we must prove $$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$ or $$x^n+y^n\ge 2$$ this last inequality seem it is not hold.so How to prove my question.Thanks

I want to see without derivative solution,such as Holder and so on — math110, Feb 03 '21 at 12:34

River Li · Answer 1 · 2021-02-04T01:32:09.687

4

Hint:

It suffices to prove that $$\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \ge \left(\frac{x^{2n+1} + y^{2n+1}}{2}\right)^{\frac{1}{2n+1}}.$$ Since this inequality is symmetric and homogeneous, assume $y = 1$ and $x \ge 1$. It suffices to prove that $$\ln (x^{n+1} + 1) - \ln (x^n + 1) \ge \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2}.$$ Take derivative.

Update

Let $f(x) = \mathrm{LHS} - \mathrm{RHS}$. We have $$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$

edited Feb 04 '21 at 01:32

answered Jan 31 '21 at 15:43

River Li

37,323

your first inequality can use Holder inequality to solve it? – math110 Feb 01 '21 at 13:21
@inequality I don't know. Can you? – River Li Feb 01 '21 at 13:28
Sorry but how is that assmption WLOG? – N.S.JOHN Feb 01 '21 at 15:39
@N.S.JOHN First, due to symmetry, we may assume $x\ge y$. Second, due to homogeneity, we may assume $y=1$. – River Li Feb 01 '21 at 16:58
Thanks. Beautiful solution – N.S.JOHN Feb 01 '21 at 17:12
Where is there homogeneity in the question? – Andreas Feb 01 '21 at 20:13
@Andreas In my first equation, you consider $(\alpha x, \alpha y)$ for $\alpha > 0$, and you will see the homogeneity. – River Li Feb 02 '21 at 05:31
Ok I see it in your equation. I was wrong to think that in the OP's question there is homogeneity. – Andreas Feb 02 '21 at 08:43
I try derivative your last inequality,it is not easy,can you complete it? Thanks – math110 Feb 02 '21 at 13:21
let $f(x)=LHS-RHS$,then $$f'(x)=\dfrac{(n+1)x^n}{x^{n+1}+1}-\dfrac{nx^{n-1}}{x^n+1}-\dfrac{x^{2n}}{x^{2n+1}+1}$$ then it's not easy to prove $f'(x)\ge 0,x\ge 1$ – math110 Feb 02 '21 at 13:28
@inequality Clearing the denominators and simplify it. – River Li Feb 02 '21 at 13:32
@inequality I added something. – River Li Feb 02 '21 at 13:34

score 3 · Accepted Answer · answered Feb 04 '21 at 10:34

Since at the point $x=y$, equality holds, and multiple derivatives are $0$, it appears likely that a solution with derivatives is needed. Here is one:

Following @River Li's calculations, it suffices to prove, for $x \ge 1$, that (let $n$ be a parameter) $$ f(x) = \ln (x^{n+1} + 1) - \ln (x^n + 1)- \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2} \ge 0 $$ This holds with equality for $x=1$, so we prove that $f(x)$ is increasing with $x$. The derivative is $$ f'(x) = \frac1x \Big[ \frac{(n+1)x^{n+1}}{x^{n+1} + 1} - \frac{nx^{n}}{x^{n} + 1}- \frac{x^{2n +1}}{x^{2n +1} + 1} \Big] $$ Multiplying out gives
$$ f'(x) = \frac{x^{n-1}}{ (x^{n+1} + 1) (x^{n} + 1) (x^{2n +1} + 1)} \Big[x - n + nx - x^{2n +1} + n x^{2n +2} - n x^{2n +1}\Big] $$ Hence it suffices to prove that the last term in brackets is $\ge 0$, slightly rewritten: $$ g(x) = x(n+1) - n + x^{2n +1}(-1 -n + n x)\ge 0 $$ Again we have equality for $x=1$. Further, note that $x(n+1) - n \ge 0$ holds always for $x \ge 1$, and $-1 -n + n x \ge 0 $ holds for $x\ge \frac{n+1}{n}$, so we only need to consider the range $x \in [1 \; \frac{n+1}{n}]$. In that range, we need to prove $$ \frac{x(n+1) - n }{1 + n - n x} \ge x^{2n +1} $$ Taking logarithms gives $$ h(x) = \ln (x(n+1) - n ) - \ln (1 + n - n x) - (2n + 1 ) \ln x \ge 0 $$ Since we have equality for $x=1$, we use the above method again and prove that $h(x)$ is increasing. Taking derivatives gives that we need to prove $$ h'(x) = \frac{n+1}{x(n+1) - n }+ \frac{n}{1 + n - n x} - (2n + 1 ) \frac1x \ge 0 $$ or, multiplying out, $$ h'(x) = \frac{1}{x(x(n+1) - n) (1 + n - n x)} \Bigg[ n(2n^2 + 3n + 1)(x - 1)^2 \Bigg] \ge 0 $$ But this is clearly true for $x \in [1 \; \frac{n+1}{n}]$, which proves the claim. $\qquad \Box$

(+1) We can also take 1st and 2nd derivatives of $F(x)=nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n$. — River Li, Feb 04 '21 at 12:38

score 1 · Answer 3 · answered Mar 14 '23 at 20:46

(Basically @River Li's idea:. I altered a bit the notations since we'll need $x$ in a different role )

One wants to show that

$$\left(\frac{t_1^{2n+1} + t_2^{2n+1}}{2}\right)^{\frac{1}{2n+1}} \le \frac{t_1^{n+1} + t_2^{n+1}}{t_1^n + t_2 ^n} $$.

since both functions are homogeneous of degree $1$.
More generally, we'll show that for all positive $0\le a< b$ the function in $x\in (\frac{b-a}{2}, a]$

$$\left(\frac{t_1^{b+x} + t_2^{b+x}}{t_1^{a-x} + t_2^{a-x}}\right)^{\frac{1}{b- a + 2 x}}$$

is decreasing.

We can choose $t_1$, $t_2$ up to proportionality, and both raised to the same power. Therefore is is enough to show that for every $a \ne b$ the function

$$\frac{f(b+x) - f(a-x)}{(b+x) - (a-x)} \ \ \ (*)$$ is decreasing in $x$, where

$$f(x) = \log \cosh x$$

Now, the derivative of the expression in $(*)$ is

$$ 2 [ (f(b+x) - f(a-x)) - \frac{1}{2} ((b+x) - (a-x)) ( f'(b+x) + f'(a-x))]$$

We recognize in the above the error for the trapezoidal rule for $$\int_{a-x}^{b+x} f'(t) \,d t$$

We know this is negative if $f'$ is concave, that is, if $$f'''<0$$

But we calculate readily $$(\log \cos h x)^{'''} = (\tanh x)^{''} = - 2 \tanh x \operatorname{sech}^2 x < 0$$ for $x> 0$.

Nice. Perhaps more details on the connection between 1st equation and 2nd equation ad 3rd equation ($(*)$) are required. — River Li, Mar 14 '23 at 23:09
@River Li: Thank you! Yes, I will have to write more details. — orangeskid, Mar 15 '23 at 02:17

ə̷̶̸͇̘̜́̍͗̂̄︣͟ · Answer 4 · 2021-02-08T15:05:13.347

A different approach is to consider the equation $x^{2n+1}+y^{2n+1}=k$ and take polar coordinates. This gives $(x,y)=(k^a\cos^{2a}\varphi,k^a\sin^{2a}\varphi)$ where $a=1/(2n+1)$. We wish to show that* if $k\ge2$ then $$\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=k^a\cdot\frac{\cos^{1+a}\varphi+\sin^{1+a}\varphi}{\cos^{1-a}\varphi+\sin^{1-a}\varphi}\ge1$$ for all $a\in(0,1)$ which we can rewrite as $$k^a\ge\sec^{2a}\varphi\cdot\frac{1+\tan^{1-a}\varphi}{1+\tan^{1+a}\varphi}=\frac{(1+s^2)^a(1+s^{1-a})}{1+s^{1+a}}$$ where $s=\tan\varphi\in[0,1]$ as the function is invariant under the transformation $s\mapsto1/s$. This means that $$k^{1-2b}\ge\frac{(1+t)^{1-2b}(1+t^a)}{(1+t^{1-a})}\implies\left(\frac{1+t}k\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\le0$$ where $t\in[0,1]$ and $b=(1-a)/2\in(0,1/2)$. Since the LHS is a decreasing function of $k$ it remains to show that $$\max_t\left\{\left(\frac{1+t}2\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\right\}=0.$$ The easiest way to prove this uses calculus and I don't think there are methods based solely on inequality results to do so.

_{*This is actually a stronger claim, that for all real $x,y,n>0$, if $x^{2n+1}+y^{2n+1}\ge2$ then $x^{n+1}+y^{n+1}\ge x^n+y^n$.}

show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$

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