Minimize $3x^2-6x+2\sqrt{x}+\sqrt{3-2x}$.

Question

WolframAlpha gives the result as follows:

$$\min\{3x^2-6x+2\sqrt{x}+\sqrt{3-2x}\}=0~~ \text{at}~~ x=1.$$

Can it be obtained by inequality without differentiation?

Answer 1

Note that $\sqrt{y} \ge \frac{2y}{1+y}$ for $y \ge 0$ (actually it is just $1+y \ge 2\sqrt{y}$).

Since $0 \le x \le \frac{3}{2}$, we have \begin{align} 3x^2 - 6x + 2\sqrt{x} + \sqrt{3-2x} &\ge 3x^2 - 6x + 2\cdot \frac{2x}{1+x} + 3\cdot \frac{2(3-2x)}{1 + (3-2x)}\\ &= \frac{3(1+x-x^2)(x-1)^2}{(1+x)(2-x)}\\ &\ge 0 \end{align} with equality if and only if $x = 1$; Here it is easy to prove that $1+x-x^2 > 0$ for $0\le x\le \frac{3}{2}$.

Answer 2

Since both $x$ and $3-2x$ are square-rooted and equal at $x=1$, it's tempting to expand around $1$ as a potential minimum; write $x=1+y$, so we want to minimize $3y^2-3+2\sqrt{1+y}+\sqrt{1-2y}$. By the binomial theorem, this expands as$$3y^2-3+2(1+\tfrac12y-\tfrac18y^2)+(1-y-\tfrac12y^2)+o(y^2)=\tfrac94y^2+o(y^2).$$Note we don't need differentiation to derive $\sqrt{1+y}=1+\tfrac12y-\tfrac18y^2+o(y^2)$, because$$(1+ay+by^2)^2=1+2ay+(a^2+2b)y^2+o(y^2)$$gives us $2a=1,\,a^2+2b=0$ to solve simultaneously.

Answer 3

We need to prove that: $$3x^2-6x+2\sqrt{x}+\sqrt{3-2x}\geq0$$ or $$3(x-1)^2+2(\sqrt{x}-1)+\sqrt{3-2x}-1\geq0$$ or $$3(x-1)^2+\frac{2(x-1)}{\sqrt{x}+1}-\frac{2(x-1)}{\sqrt{3-2x}+1}\geq0$$ or $$3(x-1)^2-\frac{6(x-1)^2}{(\sqrt{x}+1)(\sqrt{3-2x}+1)(\sqrt{3-2x}+\sqrt{x})}\geq0,$$ for which it's enough to prove that $$(\sqrt{x}+1)(\sqrt{3-2x}+1)(\sqrt{3-2x}+\sqrt{x})\geq2.$$ Can you end it now?

For example:

Let $\sqrt{x}=a$ and $\sqrt{3-2x}=b$.

Thus, $2a^2+b^2=3$, $a^2\leq\frac{3}{2}$ and $$(\sqrt{x}+1)(\sqrt{3-2x}+1)(\sqrt{3-2x}+\sqrt{x})=$$ $$=(a+1)(b+1)(a+b)=ab(a+b)+(a+b)^2+a+b\geq $$ $$\geq a^2+b^2+\sqrt{a^2+b^2}=3-a^2+\sqrt{3-a^2}\geq3-\frac{3}{2}+\sqrt{3-\frac{3}{2}}>2.$$

Answer 4

Adding the inequalities from Lemmas 1 and 2 below together yields $$ \sqrt{3-2x} + 2\sqrt x \ge x(3-2x) + 3x-x^2 = 6x-3x^2 $$ for $0\le x\le\frac32$, which is the statement to be proved. Hence it suffices to prove those two lemmas.

Lemma 1: $\sqrt{3-2x} \ge x(3-2x)$ for $0\le x\le\frac32$.

(The quadratic polynomial on the right-hand size was chosen to have the same values as $\sqrt{3-2x}$ at $x=\frac32$ and $x=1$ and also the same slope at $x=1$. Once it was chosen, the algebra magically worked out.)

Proof: When $x\ge0$ we have $0 \le (x+1)^2(2x+1) = 1-x^2(3-2x)$, and so when $0\le x\le\frac32$ we have $$ 1 \ge x^2(3-2x) \implies 1 \ge x\sqrt{3-2x} \implies \sqrt{3-2x} \ge x(3-2x). $$

Lemma 2: $2\sqrt x \ge 3x-x^2$ for $0\le x\le\frac32$.

(The quadratic polynomial on the right-hand size was chosen to have the same values as $2\sqrt x$ at $x=0$ and $x=1$ and also the same slope at $x=1$. Once it was chosen, the algebra magically worked out.)

Proof: When $x\le 4$ we have $0 \le (4-x) (x-1)^2 x = 4x - x^2(3-x)^2$, and so when $0\le x\le3$ we have $$ 4x \ge x^2(3-x)^2 \implies 2\sqrt x \ge x(3-x). $$

Answer 5

We have: $\sqrt{x} \geq \sqrt{3-2x}$ iff $x \geq 1$. We now split the interval into $[0,1]$ and $[1,3/2]$.

For $x \in [0,1]$, the given expression is at least $3(x^2-2x+\sqrt{3-2x}) \geq 3(x^2-2x+1) \geq 0$, with minimum achieved at $x=1$.

For $x \in [1,3/2]$, the given expression is at least $3(x^2-2x+\sqrt{x}) \geq 3(x^2-2x+1) \geq 0$, and once again the minimum is at $x=1$.