Solution 1 (algebraic solution due to Christopher Duff).
Using the given inequality, and then the first two equalities, we have
144
>= (xu+yv)^2
= (x^2)(u^2)+2xyuv+(y^2)(v^2)
= (x^2)(u^2)+(x^2)(v^2)+(y^2)(u^2)+(y^2)(v^2)-(x^2)(v^2)+2xyuv-(y^2)(u^2)
= (x^2+y^2)(u^2+v^2)-(xv-yu)^2
= 9*16-(xv-yu)^2,
(xv-yu)^2 <= 0,
xv-yu = 0,
(1) xv = yu.
Using (1),
(x+v)^2
= x^2+v^2+2xv
= (9-y^2)+(16-u^2)+2xv
= 25 -y^2+2yu-u^2
= 25-(y-u)^2.
Therefore x+v is maximum when y = u, and the maximum value is sqrt(25-0^2) = 5.
Solution 2 (geometric solution).
By the Cauchy-Schwarz inequality,
xu+yv <= sqrt(x^2+y^2)*sqrt(u^2+v^2) = sqrt(9*16) = 12.
Since we are given that xu+yv >= 12, so
12 <= xu+yv <= 12,
or
(1) xu+yv = 12.
This equality occurs iff there exists a number k such that
(2) u/x = v/y = k.
By (1), (2), and the first given equation,
xu+yv = 12,
x(kx)+y(ky) = 12,
k(x^2+y^2) = 12,
k*9 = 12,
k = 4/3.
Geometrically, the maximum value of x+v = x+ky = x+4y/3 occurs at the tangent point between the line whose slope is -3/4 and the circle x^2+y^2 = 9 in the first quadrant. The line of radius which intersects this tangent point has the equation y = 4x/3. Therefore the tangent point (a, b) satisfies
(1) a^2+b^2 = 9;
(2) b = 4a/3.
Solving (1) and (2),
a^2+(4a/3)^2 = 9,
25a^2/9 = 9,
a = 9/5.
It follows that b = 4a/3 = (4/3)(9/5) = 12/5 and max{x+v} = a+4b/3 = 9/5+(4/3)(12/5) = 5.
Solution 3 (tigonometric idea from Anirud Thyagharajan).
There exists angles A and B such that x = 3*cos(A), y = 3*sin(A), u = 4*cos(B), and v = 4*Sin(A), where 0 <= A, B < 2*pi. Since
xu+yv >= 12,
12*cos(A)*cos(B)+12*sin(A)*sin(B) >= 12,
cos(A-B) >= 1,
(1) A = B.
Now x+v = 3*cos(A)+4*sin(A). Let C be an angle such that cos(C) = 4/5, sin(C) = 3/5, and 0 <= C < 2*pi. Thus
x+v
= 5*[(3/5)*cos(A)+(4/5)*sin(A)]
= 5*[cos(A)*cos(C)+sin(A)*sin(C)]
= 5*cos(A-C).
It follows that x+v is maximum when A = C, and the maximum value is 5.