Prove that for any convex function $f(x)+f(y)\le f(x-1)+f(y+1)$ for all $x<y$

As per the definition of convex function
$$f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2) \forall\lambda\in[0,1]\\ \forall x_1,x_2\in\Bbb{R}$$
There is a hint for this problem- try to replace $$\lambda, x_1,x_2$$ with something involving $$x,y$$ to get that desired inequality. Remember $$x
I have also tried using slop condition of convex function, but can't prove it.
Can anyone solve this? Thanks for assistance in advance.

• I think the inequality should be $f(x) - f(y) \leq f(x-1) - f(y-1)$, otherwise there exists a counterexample. – Seewoo Lee Sep 10 at 5:00
• It is still false. – Zhaohui Du Sep 10 at 5:02
• I suspect the inequality is $f(x)+f(y) \le f(x-1) + f(y\color{red}{+}1)$ for $x < y$. If that is the case, it is a special case of Karamata's inequality and above wiki entry has a proof of that. – achille hui Sep 10 at 5:04
• @ZhaohuiDu Could you check my answer? – Seewoo Lee Sep 10 at 5:07
• @SeewooLee you modification is also true, it is also a special case of karamata inequality. – achille hui Sep 10 at 5:12

This is false for $$f(x)=x^{2}$$.

Hint for the revised question:

$$x=\alpha (x-1)+(1-\alpha) (y+1)$$ where $$\alpha =\frac {y+1-x} {y+2-x}$$. Apply the definition of convexity. Similarly we can write $$y$$ in the form $$\beta (x-1)+(1-\beta) (y+1)$$. Apply the definition again and add the two inequalities.

• There was a mistake in my question, I have fixed it now. – Biswarup Saha Sep 10 at 5:25
• @BiswarupSaha I have edited my answer accordingly. – Kavi Rama Murthy Sep 10 at 5:30
• Got it. Nice idea, thank you. – Biswarup Saha Sep 10 at 6:45

I'll give a sketch of proof for a modified problem: $$f(x) - f(y) \leq f(x-1) - f(y-1)$$. This is equivalent to $$f(x) - f(x-1) \leq f(y) - f(y-1)$$.

Convexity of $$f$$ is equivalent to the following: for $$x_{1} < x_{2} < x_{3}$$, $$s(x_{1}, x_{2}) \leq s(x_{1}, x_{3}) \leq s(x_{2}, x_{3})$$ where $$s(x, y) = \frac{f(y) - f(x)}{y-x}$$ is a slope of the secant line. Using this, we have $$f(x) - f(x-1) = s(x-1, x) \leq s(x-1, y) \leq s(y-1, y) = f(y) - f(y-1).$$ Equivalence can be proved by replacing $$\lambda, x_{1}, x_{2}$$ with appropriate variables, which may be in somewhere on Google. (I can't find it now, but I'm sure that it is true.)