Posts Tagged ‘Triangle Inequality’

A Small Substitution. Goes a Long Way!

Now, many of us (including me) are really troubled by some geometric inequalities which seem more difficult than many algebraic inequalities. But actually, as we shall see later, many of them are even weaker than AM-GM inequality. Now, many questions arise when I say this, but I hope those will be clarified as you read further.
Now, I start with some well known trigonometrical identities.

sin \ A = \frac {2 tan \ A/2}{1 + tan^2A/2}
cos \ A = \frac {1 - tan^2A/2}{1 + tan^2A/2}
Also, in a triangle ABC,
\sum_{cyc} tan \frac {A}{2} tan\frac {B}{2} = 1
Now, here comes our main substitution
In a triangle, Let tan\frac {A}{2} = x,tan\frac {B}{2} = y,tan\frac C 2 = z
So, we have xy + yz + zx = 1.
Substituting these in our earlier identities, we get
sin \ A = \frac {2x}{1 + x^2} = \frac {2x}{x^2 + xy + yz + zx} = \frac {2x}{(x + y)(x + z)}
cos \ A = \frac {1 - x^2}{1 + x^2} = \frac {1 - x^2}{x^2 + xy + yz + zx} = \frac {1 - x^2}{(x + y)(x + z)}
These little ideas really do go a very long way in proving many geometric inequalities.
We begin with the Hadwiger-Finsler Inequality.

\boxed{1} In a triangle ABC,

\boxed{2ab + 2bc + 2ca - a^2 - b^2 - c^2 \ge 4\sqrt {3}\Delta}
\underline{Proof}
By the extended law of sines, we have
a = 2Rsin \ A, \ b = 2Rsin \ B, \ c = 2Rsin \ C
Also, \Delta = \frac {abc}{4R} = 2R^2 \prod_{cyc} sin \ A
Now, putting these values in our inequality, we get
\sum_{cyc}(8R^2sin \ Asin \ B - 4R^2sin^2A) \ge 8\sqrt {3}R^2\prod_{cyc}sin \ A
\iff \sum_{cyc}(2sin \ Asin \ B - sin^2A) \ge 2\sqrt {3}\prod_{cyc}sin \ A
HERE COMES THE SUBSTITUTION!Performing our substitution, we get
\iff \sum_{cyc}(\frac {8xy}{(x + y)^2(y + z)(z + x)} - \frac {4x^2}{(x + y)^2(y + z)^2}) \ge \frac {16\sqrt {3}xyz}{\prod(x +...
\iff \sum_{cyc}(8xy(1 + z^2) - 4x^2{y + z)^2) \ge 16\sqrt3xyz
\iff 2 - \sum_{cyc}2x^2y^2 \ge 4\sqrt3xyz
Now, \sum_{cyc}x^2y^2 = (xy + yz + zx)^2 - 2xyz(x + y + z) = 1 - 2xyz(x + y + z)
\iff 2 - 2 + 4xyz(x + y + z) \ge 4\sqrt3xyz
\iff x + y + z \ge \sqrt3
which is trivial and follows from (x + y + z)^2 \ge 3(xy + yz + zx)
:wink:
Now we solve another geometric inequality by the same method, but we have to do some work before employing it. This inequality was proposed to me by Aravind Srinivas L.

\boxed{2} Given triangle ABC with cevians AD,BE,CF not necessarily concurrent, D,E,F lie on the sides BC,CA,AB respectively. Suppose AD,BE,CF are extended to meet the circumcircle of triangle ABC at P,Q,R respectively. Prove that:

\frac {AD}{DP} + \frac {BE}{EQ} + \frac {CF}{FR} \ge 9
\underline{Proof}
Note that \frac {AD}{DP} = \frac {[ADC]}{[CDP]} = \frac {[ADB]}{[BDP]} = \frac {[ADC] + [ADB]}{[CDP] + [BDP]} = \frac {[ABC]}{[CBP]}, where [\cdot]denotes area.
Therefore, we only have to prove,
\frac1{[ARB]} + \frac {1}{[BPC]} + \frac {1}{[CQA]} \ge \frac {9}{[ABC]}
Now, it is well known that area of a triangle maximizes (given that it is on a fixed base and circle and the third point lies on a particular arc i.e. major arc or minor arc) when the third point is the midpoint of the arc.

Therefore, it suffices to prove the given statement when P,Q,R are the midpoints of the arcs(minor) CB,AC,AB respectively. This happens if and only if AD,BE,CF are the angle bisectors of \triangle ABC.

Therefore, \angle PBC = \angle PAC = A/2 = \angle PAB = \angle PCB
Also, \angle BPC = 180 - A.
Notice also that \Delta ABC,\Delta BPC,\Delta CQA,\Delta ARB have the same circumradius. Let it be R.

Therefore, since area of any \Delta = 2R^2 \cdot sinA \cdot sinB \cdot sinC
\Rightarrow [ABC] = 2R^2 \cdot sinA \cdot sinB \cdot sinC
\Rightarrow [BPC] = 2R^2 \cdot sinA \cdot sin^2A/2
and similarly for others.
Thus our inequality becomes ready to be killed with our substitution. It becomes,

\sum_{cyc}\frac {1}{2R^2 \cdot sinA \cdot sin^2A/2} \ge \frac {9}{2R^2 \cdot sinA \cdot sinB \cdot sinC}
\iff \sum_{cyc} \frac {sinB \cdot sinC}{sin^2A/2} \ge 9
Thus, we can apply our substitution now. (if you don’t know how to go about the half angle sines then use the formula sin^2A/2 = (1 - cosA)/2)
Employing our substitution, we get
\sum_{cyc} \frac {\frac {4yz}{(y + z) \prod (x + y)}}{\frac {x^2}{(x + y)(x + z)}} \ge 9
\iff \sum_{cyc} \frac {4yz}{x^2(y + z)^2} \ge 9
Now, taking xy = p,yz = q,zx = r, this turns into
For positive reals p,q,r with p + q + r = 1,
\sum_{cyc} \frac {p}{(q + r)^2} \ge \frac 9 4
which is an easy application of Cauchy-Schwarz Inequality.
Indeed,
\sum_{cyc} \frac {p}{(q + r)^2} = (\sum_{cyc} \frac {p}{(q + r)^2})(p + q + r) \ge (\sum_{cyc} \frac {p}{q + r})^2\ge \frac 9...
due to Cauchy-Schwarz and Nesbitt’s inequality.
(Alternately you may try doing it by Jensen’s too!)
So…we are done :D
Now we do another problem. More harder. But easy with this method. ;)

(Source: Vietnam TST 1993 for the 34nd IMO, problem 4)
\boxed{3} In a triangle ABC, if O,H,I are the orthocentre, circumcentre, incentre respectively, then prove that

2 \cdot OI \ge IH
\underline{Proof}
2 \cdot OI \ge IH \iff 4 \cdot OI^2 \ge IH^2 \iff 4R^2 - 8Rr \ge 2r^2 - 4R^2 \cos A \cos B \cos C
(where R is the circumradius, r is the inradius)
Now, rearranging and dividing throughout by R^2 we get,
a^2 + 4a - 2 \cos A \cos B \cos C \le 2 (where a = r/R)
Since it is well known that a = 4 \prod \sin A/2, now we can apply our substitution ! :D
Now, Let x = \tan A/2 and so on to get
a = \frac {4xyz}{\prod_{cyc}(x + y)}
Since we also know how to substitute for \cos A, so our inequality becomes,
\frac {16x^2y^2z^2}{\prod_{cyc}(x + y)^2} + \frac {16}{\prod_{cyc}(x + y)} - \frac {(1 - x^2)(1 - y^2)(1 - z^2)}{\prod_{cyc}(...
\iff 8x^2y^2z^2 + 8xyz(\sum_{cyc}x - xyz) - 1 + \sum_{cyc} x^2 - \sum_{cyc} x^2y^2 + x^2y^2z^2 \le (x + y + z - xyz)^2
Note that we’ve used the identity: (x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) - xyz. Now we also use \sum_{cyc} x^2y^2 = 1 - 2xyz(x + y + z)
\iff 8xyz(x + y + z) - 1 + \sum_{cyc} x^2 - 1 + 2xyz(x + y + z) + x^2y^2z^2 \le \sum_{cyc}x^2 + 2 + x^2y^2z^2 - 2xyz(x + y + ...
\iff 12xyz(x + y + z) \le 4
Which holds by Newton’s Inequality. \rightarrow 3xyz(x + y + z) \le (xy + yz + zx)^2.
We are done! :)

The last question I’m gonna do is not an inequality but an identity. This shows that we can prove trigonometrical identities also with it.
\boxed{4}\triangle ABC is acute and h_a,h_b,h_c denote its altitudes. R,r,r_0 denote the radii of its circumcircle, incircle and incircle of its orthic triangle (whose vertices are the feet of its altitudes). Prove the relation:

h_a + h_b + h_c = 2R + 4r + r_0 + \frac {r^2}{R}
\underline{Solution}
h_a + h_b + h_c = 2R + 4r + r_0 + \frac {r^2}{R}
\iff AH + BH + CH + HD + HE + HF = 2R + 2r + r_0 + \frac {r^2}{R}
where D,E,F are the feet of altitudes.
Now, using AH + BH + CH = 2(R + r), we get
AH + BH + CH + HD + HE + HF = 2R + 2r + r_0 + \frac {r^2}{R} , \iff HD + HE + HF = 2r + r_0 + \frac {r^2}{R}
Now, using the well known formulas, HD = 2R \cos A, HE = 2R \cos B, HF = 2R \cos C, r = 4R \sin A/2 \sin B/2 \sin C/2 and r_0 = 2R \cos A \cos B \cos C, our equation reduces to
\sum_{cyc}\cos A \cos B = 4 \prod_{cyc}\sin A/2 + \prod_{cyc} \cos A + 8 \prod_{cyc}\sin^2 A/2
Now, let \tan A/2 = x , \tan B/2 = y , \tan C/2 = z
Therefore xy + yz + zx = 1, and
\cos A = \frac {1 - x^2}{(x + y)(x + z)}
\sin A/2 = \frac {x}{\sqrt {(x + y)(x + z)}}
Plugging these into the equation, bringing it to a common denominator, it reduces to
\frac {\sum(1 - x^2)(1 - y^2)(1 + z^2)}{\prod (x + y)^2} = \frac {4xyz(x + y + z) + \prod(1 - x^2) + 8x^2y^2z^2}{\prod (x + y...
After some (not SO tedious) expansion(Actually, I had written it, but it wasn’t coming out fully when I previewed it :wink: )…..as many terms cancel out…we get,
\iff 2\sum_{cyc}x^2y^2 = 2 - 4xyz(x + y + z)
\iff (xy + yz + zx)^2 = 1
Which is true.
So, we are done! :)

So….we can see how this opens up various new possibilities….
One can simply pick up a strong geometric inequality in which this inequality can be employed….and then make this substitution to turn it into an algebraic inequality….

 

Ok, So, here are some exercises to make you further in touch with this useful method.

\boxed{1} Prove that in triangle \triangle ABC
1 + \cos A \cos B \cos C \ge \sqrt {3}\sin A \sin B \sin C
\boxed{2} In triangle \triangle ABC, let D,E,F be the points where the incircle touches the sides. Let r,R,s be the inradius, circumradius, semiperimeter, respectively, of the triangle. Prove that
\frac {2rs}{R} \le DE + EF + FD \le s
and determine when equality occurs.
\boxed{3}(MOP 1998) Let \triangle ABC be an acute triangle with circumcentre O, orthocentre H and circumradius R. Show that for any point P on the segment \overline{OH},

PA + PB + PC \le 3R

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