Showing that $tf(x) + (1-t)f(y) \leq f(tx + (1-t)y)$ 
Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a twice continuously differentiable function such that $f''(x) \leq 0$. Prove that $$tf(x) + (1-t)f(y) \leq f(tx + (1-t)y)$$
  for any two points $x,y \in \mathbb{R}$ and $0\leq t\leq 1$.

I started by attempting to take the second derivative of both sides, but this just gives me an incredibly messy result.
Any input greatly appreciated.
 A: First, we note that if $f''(x)\le 0$, then for any $x$ and $y$ such that $x<y$ the fundamental theorem of calculus guarantees that
$$\begin{align}
f'(y)-f'(x)&=\int_x^y f''(t)\,dt\\\\
&\le 0
\end{align}$$
Thus, $f'$ is non-decreasing.  Therefore, by the fundamental theorem of calculus we have for $x<y$
$$\begin{align}
f(y)-f(x)&=\int_x^y f'(t)\,dt\\\\
&\le f'(x)(y-x)
\end{align}$$
Now, let $t\in (0,1)$ and let $\xi = tx+(1-t)y$.  Then, $x-\xi = (1-t)(x-y)$ and $y-\xi = t(y-x)$ and we have
$$\begin{align}
f(x)-f(\xi)&\le f'(\xi)(x-\xi) =(1-t)f'(\xi)(x-y)\tag 1\\\\
f(y)-f(\xi)&\le f'(\xi)(y-\xi) =tf'(\xi)(y-x)\tag 2
\end{align}$$
Multiplying $(1)$ by $t$ and $(2)$ by $(1-t)$ and adding yields
$$\begin{align}
tf(x)+(1-t)f(y) &\le f(\xi)\\\\
&=f(tx+(1-t)y)
\end{align}$$
And we are done!
A: Here is a more geometric, calculation light approach to the contrapositive statement, namely that if the inequality does not hold, then the second derivative must be positive somewhere.
What does the inequality mean? It means that if you draw the line segment between the points $(x, f(x))$ and $(y, f(y))$, then the graph of the function stays above that line segment. Thus we're exploring what happens if the graph is ever below the segment. Say that happens at a point $z \in (x, y)$.
Since the graph intersects the segment at $(x, f(x))$, in order for the graph to ever go below the line, the derivative of the function must be less than the slope of the line at some point between $x$ and $z$ (says the mean value theorem). Similarily, for the graph to meet up with the line segment at $(y, f(y))$, the derivative of the function must be greater than the slope of the line segment at some point between $z$ and $y$. Thus the derivative of the function has increased, which means that the second derivative must be positive somewhere.
A: Consider $a < x < b$ in $\mathbb R$. 
Let $a, a + h  \in \mathbb R$. By the Mean Value Theorem there exists $c \in (a,a + h)$ such that $$f(a)  = f(a + h) + f'(a) \cdot h + \frac{f''(c)}{2} \cdot h^2 $$
as $f'' \leq 0$ 
$$f(a + h) \leq f(a)+ f'(a) \cdot h$$
Thus $$\frac{f(a+h) - f(a)}{h} \leq f'(a) \,\,\, \text{when}\,\, h> 0 \,\, \text{and} \,\, \frac{f(a+h) - f(a)}{h} \geq f'(a) \,\,\, \text{when}\,\, h< 0  $$
Equivalently for $a < x < b$
$$\frac{f(b) - f(x)}{b-x} \leq \frac{f(x) - f(a)}{x- a} \implies (f(b) - f(x))(x -a ) \leq (f(x) - f(a))(b - x)$$
Adding $(f(x) - f(a))(x - a)$ on both sides of the last inequality yields 
$$\frac{f(b) - f(a)}{b-a} \leq \frac{f(x) - f(a)}{x- a}$$
which gives us the concavity of $f$. 
Note: Your statement is equivalent of saying that $f$ is concave. 
A: $t\in \{0,1\}$ or $x=y$ are trivial cases. For the non-trivial case , suppose, by contradiction, that  $x<y$ and $0<t<1$ and $f(tx+(1-t)y)>t f(x)+(1-t)f(y).$ 
Let $z=tx+(1-t)y.$ Observe that $\frac {y-z}{y-x}=t$ and $\frac {z-x}{y-x}=1-t.$
By the MVT (Mean Value Theorem) there exist $z_1\in (x,z)$ and $z_2\in (z,y)$ with $$f'(z_1)=(f(z)-f(x))/(z-x)\; \text { and }\;  f'(z_2)=(f(y)-f(z))/(y-z).$$ We have  $$f'(z_1)>f'(z_2)$$ $$\text { because }\quad f'(z_1)>f'(z_2)\iff  (y-z)(f(z)-f(x))>(z-x)(f(y)-f(x))\iff$$ $$\iff f(z)(y-x)>f(x)(y-z)+f(y)(z-x)\iff$$ $$\iff f(z)>f(x)\frac {y-z}{y-x}+f(y)\frac {z-x}{y-x}\iff$$ $$\iff f(z)>t f(x)+(1-t) f(y) .$$ Applying the MVT to $f'$ , there exists $z_3 \in (z_1,z_2)$ with $$f''(z_3)=\frac {f'(z_2)-f'(z_1)}{z_2-z_1}<0.$$  
Remarks. The calculations to get $f'(z_1)>f'(z_2)$ were guaranteed by considering the diagram : The point $(z,f(z))$ lies above the line-segment from  $(x,f(x))$ to $(y,f(y))$, so the slope of the line-segment from $(x,f(x)$ to $(z,f(z)), $ which is $f'(z_1),$ exceeds the slope $f'(z_2)$ of the segment from $(z,f(z))$ to $ (y,f(y)).$ .....Also, the main result holds if $f''$ exists and is never negative. The MVT holds as long as the derivative exists, so applying the MVT to $f'$ is valid whether $f''$ is or is not continuous.
