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.
Also, in a triangle ,
Now, here comes our main substitution
In a triangle, Let
So, we have .
Substituting these in our earlier identities, we get
These little ideas really do go a very long way in proving many geometric inequalities.
We begin with the Hadwiger-Finsler Inequality.
By the extended law of sines, we have
Now, putting these values in our inequality, we get
HERE COMES THE SUBSTITUTION!Performing our substitution, we get
which is trivial and follows from
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.
Note that , where denotes area.
Therefore, we only have to prove,
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.
Thus, we can apply our substitution now. (if you don’t know how to go about the half angle sines then use the formula )
Employing our substitution, we get
Now, taking , this turns into
For positive reals with ,
which is an easy application of Cauchy-Schwarz Inequality.
due to Cauchy-Schwarz and Nesbitt’s inequality.
(Alternately you may try doing it by Jensen’s too!)
So…we are done
Now we do another problem. More harder. But easy with this method.
(where is the circumradius, is the inradius)
Now, rearranging and dividing throughout by we get,
Since it is well known that , now we can apply our substitution !
Now, and so on to get
Since we also know how to substitute for , so our inequality becomes,
Note that we’ve used the identity: . Now we also use
Which holds by Newton’s Inequality. .
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.
is acute and denote its altitudes. denote the radii of its circumcircle, incircle and incircle of its orthic triangle (whose vertices are the feet of its altitudes). Prove the relation:
where are the feet of altitudes.
Now, using , we get
Now, using the well known formulas, , , , and , our equation reduces to
Therefore , and
Plugging these into the equation, bringing it to a common denominator, it reduces to
After some (not SO tedious) expansion(Actually, I had written it, but it wasn’t coming out fully when I previewed it )…..as many terms cancel out…we get,
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.
Prove that in triangle
In triangle , let be the points where the incircle touches the sides. Let be the inradius, circumradius, semiperimeter, respectively, of the triangle. Prove that
and determine when equality occurs.
(MOP 1998) Let be an acute triangle with circumcentre , orthocentre and circumradius . Show that for any point on the segment ,