How to show Young's inequality without using concavity of the logarithm? I'm currently working through a proof for Young's inequality in the simplest case:

If $a,b > 0$  , then $a \cdot b \leq \frac{a^p}{p} + \frac{b^q}{q}$ for
  $p,q$ conjugates.

The proof is relatively simple, but relies on the fact that
$$
\log\bigl(\gamma\cdot x+(1-\gamma)\cdot y\bigr) \geq \gamma\cdot \log(x) + (1-\gamma)\cdot \log(y) 
$$
for $0 < \gamma \leq 1$. I've tried proving this myself with no success, as well as searching online. Any suggestions would be appreciated.
 A: If the convexity of the exponential function is allowed, then
$$
A^{1/p}B^{1/q}=\exp\left(\frac1p\log A+\frac1q\log B\right)
\leqslant\frac1p e^{\log A}+\frac1q e^{\log B}=\frac{A}{p}+\frac{B}{q}
$$
Now let $a=A^{1/p}$ and $b=B^{1/q}$.

Alternatively, consider the function $f(t)=t^\alpha-\alpha t$ on $(0,\infty)$ with fixed $\alpha\in(0,1)$. Simple calculus gives
$$
f(t)\leqslant f(1)=1-\alpha,\quad\forall t>0
$$
which implies that
$$
t^\alpha\leqslant \alpha t+(1-\alpha),\quad\forall t>0.
$$
Now let $t=\frac{A}{B}$ and $\alpha=1/p$. Some algebra gives
$$
A^{1/p}B^{1/q}\leqslant \frac{A}{p}+\frac{B}{q}.
$$
A: If $p>0$, $q>0$ and $\frac{1}{p}+\frac{1}{q}=1$ then our inequality it's just AM-GM:
$$\frac{a^p}{p}+\frac{a^q}{q}=\frac{1}{p}\cdot{a^p}+\frac{1}{q}\cdot{b^q}\geq\left(a^p\right)^{\frac{1}{p}}\left(b^q\right)^{\frac{1}{q}}=ab$$
A: This inequality is famous and an easy consequence of Mean Value Theorem. We have $a, b, p, q$ as positive and $1/p+1/q=1$. Let $\alpha=1/p,\beta=1/q$ and replace $a^p, b^{q} $ by $a, b$ to get the inequality in the form $$a^{\alpha} b^{\beta} \leq\alpha a+\beta b$$ where $\alpha +\beta=1$. There is equality if $a=b$ and hence let $a<b$ (for $a>b$ we can interchange the roles of $a, b$).
Consider $f(x) =x^{\beta}$ so that $f'(x) = \beta x^{-\alpha} $ and by mean value theorem we can see that $$b^{\beta} - a^{\beta} =(b-a) \beta c^{-\alpha} $$ for some $c\in(a, b) $. And since $c^{-\alpha} <a^{-\alpha} $,  it follows that $$b^{\beta} - a^{\beta} <\beta(b-a) a^{-\alpha} $$ or $$a^{\alpha} b^{\beta} <\beta(b-a) +a=\alpha a+\beta b$$ Also note that if $\alpha, \beta$ are rational then there is no need of calculus and the inequality is an immediate consequence of AM-GM inequality. 
A: Here is one approach, based on the definition $$\log x = \int_{t=1}^x \frac{1}{t} \, dt.$$  Assume $0 < x < y$, so that in particular we have some $\delta > 0$ satisfying $y = x + \delta$.  We find $$\lambda x + (1-\lambda) y = x + (1-\lambda)\delta, \quad 0 \le \lambda \le 1, \quad \delta > 0.$$  Then $$\log(\lambda x + (1-\lambda)y) = \log(x + (1-\lambda)\delta) = \log x + \int_{t=x}^{x + (1-\lambda)\delta} \frac{1}{t} \, dt.$$
Next consider $$\lambda \log x + (1-\lambda) \log y = \lambda \int_{t=1}^x \frac{1}{t} \, dt + (1-\lambda) \left( \int_{t=1}^x \frac{1}{t} \, dt + \int_{t=x}^{x+\delta} \frac{1}{t} \, dt\right) \\ = \log x + (1-\lambda) \int_{t=x}^{x+\delta} \frac{1}{t} \, dt.$$  We are interested in the difference between the first and second expressions; namely that it is nonnegative:  $$\Delta(x,\lambda,\delta) = \int_{t=x}^{x+(1-\lambda)\delta} \frac{1}{t} \, dt - (1-\lambda) \int_{t=x}^{x+\delta} \frac{1}{t} \, dt.$$  With the substitution $u = t+x$ we can simplify this to $$\Delta(x,\lambda,\delta) = \int_{u=0}^{(1-\lambda)\delta} \frac{1}{u+x} \, du - (1-\lambda) \int_{u=0}^{\delta} \frac{1}{u+x} \, du.$$  Now we scale the first integrand with the transformation $$v = u/(1-\lambda), \quad du = (1-\lambda) dv,$$ giving $$\Delta(x,\lambda,\delta) = \int_{v=0}^\delta \frac{1-\lambda}{(1-\lambda)v + x} \, dv - \int_{u=0}^\delta \frac{1-\lambda}{u+x} \, du = (1-\lambda) \int_{t=0}^\delta \left(\frac{1}{(1-\lambda)t + x} - \frac{1}{t+x} \right) \, dt.$$  But since $1-\lambda > 0$, $\delta > 0$, and the integrand is clearly nonnegative since $$(1-\lambda) t \le t \\ \iff (1-\lambda)t + x \le t + x \\ \iff \frac{1}{(1-\lambda)t + x} \ge \frac{1}{t+x},$$ the claim is proved.
A: I will prove the following statement. Let $f(x)$ be a twice-differentiable function on $I = [a, b]$ such that for every $x \in I$, $f''(x) \leq 0$. Then for any $t \in [0, 1]$ and for any $c, d \in I$, $f(tc + (1-t)d) \geq tf(c) + (1-t)f(d)$.
Proof: Let $c, d \in I$. Let $g(x) = f(x) - \left( \frac{f(d)-f(c)}{d-c}(x-c) + f(c) \right)$. Then $g(c) = g(d) = 0$ and $g''(x) = f''(x) \leq 0$ for all $x \in I$.
We claim that $g(x) \geq 0$ for all $x \in [c, d]$. For the sake of contradiction, suppose that there is some $x \in [c, d]$ such that $g(x) < 0$. Then
$$ 0 > g(x) = \int_c^x g'(s) ds + g(c) = \int_c^x g'(s) ds .$$
This means that for some $y \in [c, x]$, $g'(y) < 0$. Then for any $z \in [y, d]$, since $g''(s) \leq 0$ for any $s \in I$,
$$ g'(z) = \int_y^z g''(s) ds + g'(y) < 0 .$$
But this implies
$$ 0 = g(d) = \int_c^d g'(s) ds  = \int_c^x g'(s) ds +  \int_x^d g'(s) ds < 0 ,$$
a contradiction. Thus $g(x) \geq 0$ for all $x \in [c, d]$.
In particular, let $x = tc + (1-t)d$. Then
$$ 0 \leq g(tc + (1-t)d) = f(tc + (1-t)d) - \left( \frac{f(d)-f(c)}{d-c}(tc + (1-t)d - c) + f(c) \right) $$
$$\rightarrow \frac{f(d) - f(c)}{d-c}(1-t)(d-c) + f(c) \leq f(tc + (1-t)d) $$
$$\rightarrow tf(c) + (1-t)f(d) \leq f(tc + (1-t)d) .$$
A: From Wikipedia's article on concave functions: "A differentiable function $f$ is concave on an interval if and only if its derivative function $f′$ is monotonically decreasing on that interval, that is, $f''<0$."
Note that $(\log(x))''=(1/x)'=-1/x^{2}<0$, so $\log$ is concave, and your inequality follows.
