a small sample of Peter Woo's solved Crux Problems

Crux Problem 4129

Peter Y. Woo, 3/04/2017

Biola University, La Mirada, Ca. 90639

        Problem 4129. (Proposed by Lorean Saccanu).

Let ABC be an acute-angled triangle and let k = 3(2-Ö3). Prove that

sec A + sec B + sec C ³ k + tan A + tan B + tan C.

        Proof. Let a=A/2, b=B/2, g=C/2.

We shall prove f(a) + f(b) + f(c) ³ k . . . (i)

where f(f) =def sec 2f - tan 2f. Let s = sinf, c = cosf, t = tanf.

        Then f(f) = (1- sin2f)/cos2f (using trig formulas like crazy)

= [(s2+c2) - 2 s c ] / (c2 -s2) = (c-s)/(c+s)

= 1/[ (1+t)/(1-t)] = cot (45°+f). (Most USA highschoolers do not know this goodie formula)

      Now a, b, g are between 0 and p/2, their average being p/6 and sum being p/2.

Hence on the xy-plane the points P(a, f(a)), P'(b, f(b)), P"(g, f(g)) lie on the curve

y = cot(45°+x) that is concave upwards.

Hence the geometric centroid of triangle PP'P" is (p/3, y'),

where y' is (1/3)(cot(45° +a) +cot(45°+b) +cot(45°+g))

and must be ³ cot(45°+p/6) = sec(30°) - tan(30°) = k.

        The whole problem is thus proved. 

Crux Problem 4197

Peter Y. Woo, 4/02/2017

Biola University, La Mirada, Ca. 90639

        Problem 4197. (Proposed by Michel Bataille)

Let x,y,z all > 0 satisfying xy +yz +zx +2xyz = 1. . .(i)

Prove that (a) 1/x +1/y +1/z ³6 and (b) x+y+z ³ 3/2.

        Proof. Let k3 = xyz.

Then xy +yz +zx £ 3k2, so that

3 k2 + 2 k3 £ 1.

Hence k £ 1/2.

        Now from (i) we get 1/x +1/y +1/z + 2 = 1/k3 which³ 8.

Hence (a) is proved.

        Again, from (i) xy + yz +zx = 1 -2k3 which ³ 1 -1/4 = 3/4.

Hence (x+y+z)2 = (x2+y2+z2)+2(xy+yz+zx) ³ 3(xy+yz+yz) which ³9/4.

Hence x+y+z ³ Ö(9/4) = 3/2, thus proving (b). (QED) 

Crux Problem 4240

Peter Y. Woo, 1/10/2018

Biola University, La Mirada, Ca. 90639

        Problem 4240. (Proposed by Rozenberg & Giugiuc.)

Let a,b,c be all > 0 with a+b+c = 1/a +1/b +1/c . . . (i)

Prove that 1+a+b+c ³ 4abc . . . . (ii)

        Proof. WLOG, assume a < b < c. We shall write a+b/c+d to mean a + (b/c) + d because the symbols / and * have stronger binding power than + and -.

      It is impossible for a,b,c to be all > 1 or all < 1. Otherwise (i) cannot be satisfied. So we have only 2 cases: Case 1: a £ b £ 1 < c or Case 2: a < 1 £ b £ c .

      Strategy: Observe 1/a+1/b +1/c = (ab + bc + ca)/abc,

hence 4abc = 4(ab+bc+ca)/(1/a + 1/b + 1/c) = 4(ab+bc+ca)/(a+b+c) . . . (iii)

On the other hand, 1 + a + b + c = (a+b+c+a2+b2+c2+2ab+2bc+2ca) / (a+b+c) . . (iv)

To prove (iii) < (iv) is to prove

a+b+c+a2+b2+c2 - 2(ab+bc+ca) > 0 . . . (v)

      There is a function we need to familiarise: y = x - 1/x = f(x) for 0 < x < ¥. It is one arm of a hyperbola, with asymtotes being

y = x and y = 0. So it is a convex curve for 0 < x < ¥, with f(1) = 0.

The requirement a + b + c = 1/a + 1/b + 1/c is equiv. to f(a) + f(b) + f(c) = 0.

In particular case, if a < b < 1 < c, then we can

replace b by 1 and c by d = 1/a such that f(b) + f(c) = f(1) + f(d),

and because of convexity of the curve y = f(x), d has to be less than c,

namely d satisfies d = 1/a so that f(d) = - f(a) and f(1) = 0,

thus satistying f(a) + f(b) + f(c) = f(a) + f(1) + f(1/a) ,

and 1 + d < b + c due to convexity. This trick shall be called Lemma 1.

Now this is case 1: LHS(v) = a + b + c + a2 +(b-c)2-2a(b+c) using Lemma1,

> a + (1 + 1/a) + a2 + (b-c)2 -2a (1+1/a)

= (b-c)2 + (1/a) [a2 + a + 1 + a3 -2a2 -2a]

= (b-c)2 + (1/a) (1-a) (1-a2) > 0, thus (v) is true.

Again for case 2: a < 1 < b < c, replace a by 1/c and b by 1

then LHS(v) = (a + b) + c + c2 +(a-b)2 -2c(a+b), using Lemma 1,

> (1+1/c) + c + c2 -2c(1+1/c) + (a-b)2

= (a-b)2 + (1/c)[ c - 1 - c2 + c3 ]

= (a-b)2 + (1/c) (c-1) (1 +c2 ) > 0, thus (v) is true.

        We have proved (v). QED 

Crux Problem 4246

Peter Y. Woo, 6/15/2017

Biola University, La Mirada, Ca. 90639

Problem. Let a+b+c+d = 1/a +1/b +1/c +1/d . . . (i)

Find the lower bound of abc+ abd+ acd + bcd.

         Proof/. Let a =b =c = e, a very small number compared with 1.

Then 1/a +1/b +1/c = 3/e, = d- 1/d + 3e.

Hence d < and very close to 3/a = 3/e.

Hence abc + abd +acd +bcd = (1/a + 1/b +1/c + 1/d) (abcd)

= (a+b+c+d) (abcd) < (3e + 3/e) e3 3/e = 9 (e3 + e)

which is arbitrarily small.

      Hence the lower bound is zero.