Minimum Value question.

Question

$\text{Problem:}$ If, $A+B+C=90°$ then find the minimum value of $13\tan^2(A)+9\tan^2(B)+\tan^2(C)$ .

My Approach: I kind of started by using and applying the $\text{AM-GM}$ inequality repeatedly but that didn't really help me out. Then I referred to the solution given in the Book, it stated that the minimum value of the given equation will be the $t\sqrt{2}$ where $t$ is the root of the equation $2t^2-45t+234=0$ . But, I wasn't able to understand how was this expression obtained. I've a feeling that it's an extensive application of the $\text{AM-GM}$ inequality, but who knows, I may be wrong. So, I request you all to help me out.

Answer 1

Hint

$$\sum_{\text{cyc}}(p\tan A-q\tan B)^2\ge0$$

Solve for $\dfrac{p^2+u^2}{13}=\cdots=m$(say)

$pq=rs=tu=n$(say) to use $$\sum_{\text{cyc}}\tan A\tan B=1$$

Answer 2

Let $a,b,c$ be the values of the tangent for $A,B,C$. The given relation $A+B+C=90^\circ $ leads to a zero value of the cotangent, so $$ 0 = \frac{1-ab-bc-ca}{a+b+c-abc}\ . $$ So we have to minimize the expression $13a^2+9b^2+c^2$, constrained by the relation $ab+bc+ca=1$. We then search for the "best $t$", so that the following relation is satisfied: $$ 13a^2+9b^2+c^2\ge 2t(ab+bc+ca)\ . $$ Because the inegquality is homogenous in $a,b,c$, we can forget a while about the constraint $ab+bc+ca=1$.

First of all, note that $t\le 3$, because we must have in particular $9b^2+c^2\ge 2t\; bc$. We have reduced the problem to finding the optimal (positive) value for $t$, so that the quadratic form corresponding to the matrix $$ M(t) = \begin{bmatrix} 13 & -t & -t \\ -t & 9 & -t \\ -t & -t & 1 \end{bmatrix} $$ takes only values $\ge 0$. Sylvester criterion. The elements on the diagonal are $>0$. The determinants of the minors $2\times 2$ built on the diagonal are also $>0$ for $t<3$. (The value $t=3$ does not work, as we see in a second.) The determinant of the matrix $M$ is $$ f(t) = \det M(t) = -2t^3 - 23t^2 + 117\ , $$ and we have for instance $f(2) = 9$, $f(3)=-144$. So we expect a root in between, rather near $2$, it is approximatively $$ \tau \approx 2.07584311418034\dots\ . $$ Sage code computing it:

sage: var('t');
sage: A(t) = matrix( 3, 3, [13,-t,-t, -t,9,-t, -t,-t,1] ).det().factor()
sage: A(t).roots(ring=RR, multiplicities=False)
[-11.0181176066186, -2.55772550756178, 2.07584311418034]

This is the optimal value, so that we have for all $a,b,c\in\Bbb R$ the inequality $$13a^2+9b^2+c^2\ge 2\tau(ab+bc+ca)\ ,$$ so the minimal value of the expression is $\boxed{\ 2\tau\approx 4.15168622836067\dots\ }$ .

$\square$

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Numerical note:

One can also use sage or an other computer algebra system to get an exact formula, but the numerical value is simpler to check, as in the sequel. (I always check and want to see the number in such situations.)

So let us check also with sage, that this is (numerically) a (local) minimal value (going always down from a chosen point):

sage: f = lambda p: 13*p[0]^2 + 9*p[1]^2 + p[2]^2 
sage: c = lambda p: p[0]*p[1] + p[1]*p[2] + p[2]*p[0] -1
sage: tau = [ root for root in A(t).roots(ring=RR, multiplicities=False) if root > 0 ][0]
sage: tau
2.07584311418034
sage: 2*tau
4.15168622836067

sage: minimize_constrained( f, [c], (0,1,1) )
(0.2780871966139785, 0.37845809756156296, 1.3628239288343866)
sage: minimize_constrained( f, [c], (1,1,0) )
(0.27811224400155166, 0.37845335977224853, 1.3627693434198571)
sage: minimize_constrained( f, [c], (1,0,1) )
(0.27805233412191027, 0.3784732000997097, 1.362878654212287)
sage: f(_)
4.15168620148928

Also, if we want to do numerical optimization for the given function mot-a-mot:

sage: g = lambda p: 13*tan(p[0])^2 + 9*tan(p[1])^2 + tan(p[2])^2 
sage: d = lambda p: p[0]+ p[1] + p[2] - pi.n()/2.
sage: m = minimize_constrained( g, [d], (pi/6.,pi/6.,pi/6.) )
sage: g(m)
4.151686307799218
sage: d(m)
0.0
sage: 2*tau
4.15168622836067

The code is rather easy to digest. Above, $f$ implements the function $f(p)=13p_0^2+9p_1^2+p_2^2$, so taking $p=(a,b,c)$ we get our function. The condition / constraint is also defined using a $p$ total variable.

Starting from $(a,b,c)=(1,1,0)$ (which satisfies $ab+bc+ca=1$) we were accompanied to a smaller value that matches numerically the value $2\tau$. Also, starting from $(\pi/6,\ \pi/6\ \pi/6)$ and using the original function we get a value that rather confirms the $2\tau$ minimum.

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Note: I suspected a typo in the original source (not here in the OP), and experimented with similar expressions, so that the corresponding matrix $M_{?,?,?}(t)$ with diagonal $13,9,1$ replaced by $?,?,?$ would lead to a negative rational root, so that we can divide by the corresponding factor and reduce the problem to "quadratic numbers". For instance for $13, 9, 9$ or for $13,9,13$. But then the $\sqrt 2$ in that $t\sqrt 2$ from the original post gave me some problems. The factor of $\det M_{?,?,?}(t) $ of second degree should have the square of its discriminant in $\Bbb Q(\sqrt 2)$, this is the only way i imagine that the hint in that book would make sense. Well, i have to stop here.