Prove $a+b+c \geq \frac{3+\sqrt{7}}{2}$ for $a,b,c \in [0, \frac32]$ with $a^2+b^2+c^2+abc=4$ [closed]

Question

Let $a,b,c \in \left[0;\frac{3}{2}\right]$ and $a^2+b^2+c^2+abc=4$. Prove that $$a+b+c \geq \dfrac{3+\sqrt{7}}{2}.$$

Source: This is a math problem that my teacher gave me $3$ months ago (the submission deadline has expired). My teacher wrote a book and sent me to test the difficulty of the problem.

My attempt: I have converted $a$ to $b,c$, used to trigonometric conversion, but all failed.

Related problem (the same source, with the same conditions): https://artofproblemsolving.com/community/c6h2975620p26673082

Please give me a suggestion! Thank you!

Answer 1

It suffices to prove that $$2a + 2b + 2c - 3 \ge \sqrt 7.$$

Using $(a + b + c)^2 \ge a^2 + b^2 + c^2$ and $(a + b + c)^3/27 \ge abc$, we have $(a + b + c)^2 + (a + b + c)^3/27 \ge 4$ which results in $a + b + c > 3/2$ or $2a + 2b + 2c - 3 > 0$.

Thus, it suffices to prove that $$(2a + 2b + 2c - 3)^2 - 7 \ge 0.$$

We have \begin{align*} &(2a + 2b + 2c - 3)^2 - 7 - 4(a^2 + b^2 + c^2 + abc - 4)\\ ={}& (-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)\\ \ge{}& 0. \tag{1} \end{align*} (Note: If $-4ab + 8a + 8b - 12 \ge 0 $, clearly (1) is true.
If $-4ab + 8a + 8b - 12 < 0$, we have
$(-4ab + 8a + 8b - 12)c + 2(3-2a)(3-2b)$
$\ge (-4ab + 8a + 8b - 12)\cdot \frac32 + 2(3-2a)(3-2b)$
$ = 2ab \ge 0$.)

We are done.

Answer 2

Just to use Lagrange's multipliers to solve this one (as the OP commented Langrange cannot find the minimum in this problem): let $\mathrm L = a^2+b^2+c^2-\lambda\,abc$, then we have the equations to solve $$2a=\lambda bc, 2b=\lambda ca, 2c=\lambda ab \implies a=b=c$$ and the constraint $a^2+b^2+c^2+abc=4 \implies a=b=c=1$ as the only candidate point in the interior $a, b, c \in (0, \frac32)$ for a minimum, and here $\implies a+b+c=3^*$.

Checking for the boundaries, we must have at least one variable zero, so WLOG let $c=0 \implies a^2+b^2=4$, and we want to minimise $a+b$, gives $\{a,b\}=\{\frac32, \frac{\sqrt7}2\}$, which gives a value of $a+b=\frac{3+\sqrt7}2<3$, so clearly this gives the minimum.

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$^*$ Note: it turns out we don't need to check if this point is actually a minimum, as we have found a lower one subsequently.

Answer 3

Rewriting the inequality as :

$$\frac{1}{2}\left(-bc+\sqrt{b^2c^2-4\left(b+c\right)^{2}+8bc+16}\right)+b+c \geq \dfrac{3+\sqrt{7}}{2}$$

We make :

$$b+c=\operatorname{constant}=C$$

The function :

$$g(bc)=-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}$$

Is increasing as $bc$ increases with $-bc+\sqrt{b^2c^2-4\left(C\right)^{2}+8bc+16}=a\geq 0$

So if $b=0$ with the constraint we have an equality as $c=1.5$

Done .