Why is this Inequality True for all Positive Real Numbers? I was reading Folland's Real Analysis and came across the following inequality in chapter 6 (it is from lemma 6.1)

Let $0<\lambda<1$. Then $t^\lambda\leq \lambda t +(1-\lambda)$, $t$ a
  real positive number.

Is this inequality always true? I'm not sure how to prove it and doesn't seem evident to me. 
 A: By the concavity of $\log$, $\lambda \log t + (1-\lambda) \log 1 \leq \log (\lambda t +(1-\lambda))$. $e^x$ is an increasing function, so 
$$\begin{align*}e^{\lambda \log t}&\leq e^{\log(\lambda t +(1-\lambda))}\\
t^\lambda &\leq \lambda t +(1-\lambda)\end{align*}$$
A: Let $f(t) = t^\lambda.$ Because $0 < \lambda < 1,$ $f''(t) < 0$ on $(0,\infty).$ So $f$ is strictly concave on $[0,\infty).$ Thus $f$ stays below any of its tangent lines, in particular the tangent line at $(1,1),$ which has the equation $ y = \lambda t + (1-\lambda).$
A: If you want to avoid concavity arguments just note that the function $$\phi(t) = t^\lambda - \lambda t - (1-\lambda)$$
has derivative $$\phi'(t) = \lambda t ^{\lambda - 1} - \lambda$$ which is positive if $0 < t < 1$ and negative if $t > 1$. Thus $\phi$ increases to its maximum at $t = 1$ and then decreases so that $$t^\lambda - \lambda t - (1-\lambda) \le \phi(1) = 0$$ for all $t > 0$.
A: It is a corollary of the general theorem of means:
General Theorem of Means: If $a_{1}, a_{2}, \ldots, a_{n}$ and $x_{1}, x_{2}, \ldots, x_{n}$ are positive real numbers with $$x_{1} + x_{2} + \cdots + x_{n} = 1$$ then $$a_{1}^{x_{1}}a_{2}^{x_{2}}\cdots a_{n}^{x_{n}} \leq x_{1}a_{1} + x_{2}a_{2} + \cdots x_{n}a_{n}\tag{1}$$ There is inequality unless all the $a_{i}$'s are equal.
The inequality in current question comes from $n = 2$, $a_{1} = t, a_{2} = 1, x_{1} = \lambda, x_{2} = 1 - \lambda$ and we get $$t^{\lambda} \leq \lambda t + 1 - \lambda\tag{2}$$ for all positive numbers $t, \lambda$ with $0 < \lambda < 1$. If $t = 1$ then there is equality otherwise there is inequality.

The general theorem of means is easily proven via induction if it is already established for $n = 2$. Hence we first establish it for $n = 2$. Changing notation for ease of typing we prove that $$a^{\alpha}b^{\beta} \leq \alpha a + \beta b\tag{3}$$ where $a, b, \alpha, \beta$ are positive and $\alpha + \beta = 1$. There is inequality unless $a = b$.
There are many ways to prove $(3)$ I provide one which is based on mean value theorem. Since $\alpha + \beta = 1$ we can write the inequality as $$a^{\alpha}b^{1 - \alpha} \leq \alpha a + (1 - \alpha) b \tag{4}$$ Clearly if $a = b$ then there is equality so let's assume $a < b$ (the case $a > b$ is same as interchanging roles of $\alpha, \beta$). The inequality $(4)$ can be written as $$b^{1 - \alpha}\leq \alpha a^{1 - \alpha} + (1 - \alpha)ba^{-\alpha}$$ or $$b^{1 - \alpha} - a^{1 - \alpha} \leq (\alpha - 1)a^{1 - \alpha} + (1 - \alpha)ba^{-\alpha}$$ or $$b^{1 - \alpha} - a^{1 - \alpha} \leq (1 - \alpha)(b - a)a^{-\alpha}\tag{5}$$ We prove that there is ienquality in $(5)$ if $0 < \alpha < 1$ and $0 < a < b$. To do so we apply mean value theorem on $f(x) = x^{1 - \alpha}$ on interval $[a, b]$ to get $$f(b) - f(a) = (b - a)f'(c)$$ or $$b^{1 - \alpha} - a^{1 - \alpha} = (b - a)(1 - \alpha)c^{-\alpha}$$ where $a < c < b$. Since $\alpha > 0$ it follows that $c^{-\alpha} < a^{-\alpha}$ and hence we get $$b^{1 - \alpha} - a^{1 - \alpha} < (1 - \alpha)(b - a)a^{-\alpha}$$ as desired. Since the current question is based on the general theorem of means for $n = 2$ the above proof is all we need to solve the current question. BTW there is another proof for $(3)$ which makes use of the fact that $f(x) = \log x$ is concave on $(0, \infty)$ and the reader should be able to complete this proof on his own.

However just to complete the story we establish the general theorem of means $(1)$ via induction. Suppose it is true for $n = 2, 3, \ldots, m - 1$ and let's try to establish it for $n = m$. Let $$x_{1} + x_{2} + \cdots + x_{m - 1} = x$$ so that $x + x_{m} = 1$ and let $$y_{1} = x_{1}/x, y_{2} = x_{2}/x, \cdots, y_{m - 1} = x_{m - 1}/x$$ so that $$y_{1} + y_{2} + \cdots + y_{m - 1} = 1$$ Then we have
\begin{align}
a_{1}^{x_{1}}a_{2}^{x_{2}}\cdots a_{m}^{x_{m}} &= (a_{1}^{y_{1}}a_{2}^{y_{2}}\cdots a_{m - 1}^{y_{m - 1}})^{x}a_{m}^{x_{m}}\notag\\
&\leq x(a_{1}^{y_{1}}a_{2}^{y_{2}}\cdots a_{m - 1}^{y_{m - 1}}) + x_{m}a_{m}\text{  (inequality (1) holds for }n = 2)\notag\\
&\leq x(y_{1}a_{1} + y_{2}a_{2} + \cdots + y_{m - 1}a_{m - 1}) + x_{m}a_{m}\text{  (inequality (1) holds for }n = m - 1)\notag\\
&= x_{1}a_{1} + x_{2}a_{2} + \cdots x_{m}a_{m}\notag
\end{align}
The general theorem of means $(1)$ is now established via induction.
