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Let $(B_t)$ the brownian motion. Why $$\mathbb E[\sup_{0\leq s\leq t}(as+cB_s)^2]\leq 2(a^2t^2+4b^2\mathbb E[B_t^2]).$$ Notcie that we have used $(a+b)^2\leq 2a^2+2b^2$. The thing is why $$\mathbb E\sup_{0\leq s\leq t}B_s^2=\mathbb E[B_t^2] \ \ ?$$

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    Think you have some typos and you should probably clarify. The last equality seems wrong. Usually when you're constraining the mean of sup as less then the mean, it's related to Doob's submartingale inequality. – 2017-01-17

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First of all: No, Brownian motion is not increasing.

Doob's maximal inequality states that for any square-integrable martingale $(M_t)_{t \geq 0}$ with continuous sample paths, we have

$$\mathbb{E} \left( \sup_{s \leq t} M_s^2 \right) \leq 4 \mathbb{E}(M_t^2), \qquad t \geq 0.$$

Since the Brownian motion $(B_t)_{t \geq 0}$ is a square-integrable martingale with continuous sample paths, this gives the estimate which was used to prove the inequality in your question.

Alternative reasoning: Using the reflection principle, it is possible to show that $$M_t :=\sup_{s \leq t} B_s \stackrel{d}{=} |B_t| \quad \text{and} \quad m_t :=- \inf_{s \leq t} B_s \stackrel{d}{=} |B_t|.$$ Since $$\sup_{s \leq t} B_s^2 \leq M_t^2 + m_t^2$$ we get $$\mathbb{E} \left( \sup_{s \leq t} B_s^2 \right) \leq 2 \mathbb{E}(B_t^2).$$