Posted on - 17-02-2017

IIT JEE

Let a_{1},
a_{2}, . .. . . . , a_{n} be n positive real numbers,
then we define their arithmatic mean (A) , geometric mean (G) and
harmonic mean (H) as A = ,

G = (a_{1} a_{2} . . . . a_{n})^{1/n} and H
= .

It
can be shown that A ≥ G ≥ H . Moreover equality holds at
either place if and only if a_{1} = a_{2} = ……. = a_{n}
.

Let a_{1}, a_{2}, .
.. . . . , a_{n} be n positive real numbers and m_{1}, m_{2},
. .. . . . , m_{n} be n positive rational numbers. Then we define
weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted
harmonic mean (H*) as .

A* = , G* = and

H* = .

It can be shown that A* ≥ G* ≥ H* . Moreover equality holds at either place if and
only

if a_{1} = a_{2} = . . . . . = a_{n} . .

If a_{i}'s and b_{i}'s are
reals, then (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})
≥ (a_{1}b_{1}
+ a_{2}b_{2} +...+ a_{n}b_{n})^{2}.

Equality holds when .

Let ABC and PQR be two given
triangles such that

sinA = , where a, b, c and p,
q, r are the sides of the triangle ABC and PQR respectively. If p, q, r(p
< q < r) are consecutive natural numbers and perimeter of the
triangle ABC is equal to 12l , then find the sides of the
triangle PQR and that of triangle ABC.

Since (a^{2}
+ b^{2} + c^{2}).( r^{2} + p^{2} + q^{2})
≥ (ar + bp + cq)^{2}
, sinA ≥ 1.

But sinA £ 1

&⇒
there is only one possibility , which is sinA = 1

&⇒ A = 90° and

&⇒ ( say
) . . . . (1)

&⇒ D
ABC and D PQR are similar
since ∠A = p/2, ∠R = p/2
(as r is the greatest side). .

In DPQR, q = p +1, r = p+2 and also r^{2}
= p^{2} + q^{2}

&⇒ (p+2)^{2} = p^{2} +
(p +1)^{2}

&⇒ p =-1, 3

&⇒ q = 4, r = 5. .

Now a = 5t, b = 3t and c = 4t from (1)

&⇒ 5t +3t + 4t = 12l

&⇒
t = l

&⇒ a = 5l , b = 3l,
c = 4l. .

Let and , where = ar + bp + cq

&⇒ cosq = ar + bp + cq , where q is the angle between .

&⇒ cos^{2}q =

&⇒ .
. . (1)

But sinA = . . . (2)

From (1) and (2) only equality will hold

i.e.

&⇒ cos^{2}q =1 and sinA =1

&⇒ q
= 0 or p and A = p/2

&⇒ are either collinear
or parallel

&⇒

&⇒ .
. . (3)

&⇒ D
ABC and D PQR are similar
since ∠A = p/2, ∠R = p/2
(as r is the greatest side). .

In DPQR, q = p +1, r = p+2 and also r^{2}
= p^{2} + q^{2}

&⇒ (p+2)^{2} = p^{2} +
(p +1)^{2}

&⇒ p =-1, 3

&⇒ q = 4, r = 5. .

Now a = 5t, b = 3t and c = 4t from (3)

&⇒ 5t +3t + 4t = 12l

&⇒
t = l

&⇒ a = 5l , b = 3l,
c = 4l. .

(i) Any inequality has to be solved using a clever manipulation of the previous results.

(ii) Any inequality involving the sides of a triangle can be reduced to an inequality involving only positive reals, which is generally easier to prove

For the triangle we have the constraints a + b > c, b + c > a, a + c > b

Do the following : put x = s- a, y = s – b, z = s - c

then, x + y + z = 3s - 2s = s and a = y +z, b = x + z, c = x + y.

Substitute a = y + z, b = x + z, c = x + y in the inequality involving a,b,c to get an inequality involving x,y, z.

Also note that the condition a + b > c is equivalent to

a + b + c > 2c i.e., 2s > 2c OR s - c > 0, i.e., z > 0.

Similarly b + c > a º x > 0, a + c > b º y > 0

\ The inequality obtained after the substitution is easier to prove. (involving only positive reals without any other constraints).

If a,b,c are the sides of a triangle and .

Prove that 8(s - a) (s - b) (s - c) £ abc

Let x = s - a, y = s - b, z = s - c, then a = y + z, b = x + z, c = x + y

then the inequality reduces to 8xyz £ (x + y)(y + z)(x + z) x,y,z ≥ 0

which follows easily from A.M. ≥ G.M. inequality.

\ (x + y) (x + z)(x + z) ≥ 8xyz. Hence proved.

Let a_{1},
a_{2}, . .. . . . , a_{n} be n positive real numbers ( not
all equal) and let m be a real number , then .

>if m ∈ R –[0, 1].

However if m∈ (0, 1), then < .

Obviously if m∈ {0, 1} , then = .

Show that x +≥ 2, if x > 0 and x +£ -2 , if x < 0 .

Since x > 0 , ≥ ( A.M. ≥ G. M. )

&⇒ x +≥ 2.

If x < 0 , let y = -x , then y > 0 and y + ≥ 2

&⇒ – x – ≥ 2

&⇒ x +£
-2.

If a_{i }’s
are all positive real numbers then prove that

(1+ a_{1} +a_{1}^{2}
) (1+ a_{2} +a_{2}^{2} ) . . . . (1+ a_{n} +a_{n}^{2}
) ≥ 3^{n} a_{1}a_{2 }….. a_{n}
.

≥ (1. a_{i} . a_{i}^{2} )^{1/3}
= a_{I} ( i = 1, 2, 3, . . . . , n)

&⇒ 1+a_{i} +a_{i}^{2}
≥ 3a_{i}

&⇒ (1+ a_{1}
+a_{1}^{2} ) (1+ a_{2} +a_{2}^{2} ) . .
. . (1+ a_{n} +a_{n}^{2} ) ≥ 3^{n} a_{1}a_{2
}….. a_{n} .

If x, y, z are positive real numbers , such that x+y+z = a , then prove that .

Since A.M. ≥ H.M. .

&⇒
.

If a, b, c are positive real numbers such that a+b+c = 18,

find the maximum
value of a^{2} b^{3}c^{4} . .

a+b+c
= 18

&⇒

&⇒ £
(theorem of weighted means)

&⇒ a^{2} b^{3} c^{4}
£ 2^{9} . 2^{2}
. 3^{3} .4^{4}.

Equality will hold, when

&⇒ a= 4, b = 6, c = 8.

Thus maximum value
of a^{2} b^{3} c^{4} is 4^{2}. 6^{3}
.8^{4} = 2^{19} .3^{3} .

If a, b, c are positive real numbers such that a+b+c = 1 , then prove that .

We have to show that

i.e

Now A.M. of mth power ≥ mth power of arithmetic mean (m= -1 here).

&⇒

&⇒ .