Q.94. Algebraic Inequality.

Question 94.
Suppose a,b,c are real numbers such that a^2 * b^2 + b^2 * c^2 + c^2 * a^2 = k, where k is a constant,
then prove that - k/2 ≤ abc(a + b + c) ≤ k.

Answer 94.
(ab + bc + ca)^2 ≥ 0
=> a^2b^2 + b^2c^2 + c^2a^2 + 2abc(a + b + c) ≥ 0
=> k + 2abc(a + b + c) ≥ 0
=> - k/2 ≤ abc(a + b + c) ... ( 1 )

(ab - bc)^2 + (bc - ca)^2 + (ca - ab)^2 ≥ 0
=> 2(a^2b^2 + b^2c^2 + c^2a^2) - 2abc(a + b + c) ≥ 0
=> k - abc(a + b + c) ≥ 0
=> abc(a + b + c) ≤ k ... ( 2 )

From equations ( 1 ) and ( 2 ),
- k/2 ≤ abc(a + b + c) ≤ k.

LINK to YA!

Q.74. Algebraic inequality

Question 74.
If a,b,c > 0 and ab + bc + ca = 1 - 2abc, then prove that  2(a + b + c) + 1 ≥ 32abc.

Answer 74.
AM ≥ HM
=> (ab + bc + ca + 2abc)/4 ≥ 4 / [1/ab + 1/bc + 1/ca + 1/2abc)] ... (1)

Now, ab + bc + ca + 2abc = 1
and 1/ab + 1/bc + 1/ca + 1/2abc = [2(a + b + c) + 1] / 2abc

Plugging in ( 1 ),
1/4 ≥ 4 / [2(a + b + c) + 1] / 2abc
=> 2(a + b + c ) + 1 ≥ 32abc.

LINK to YA!

Q.36. Inequality of Trigonometry

Question 36.
Let a, b and c be the sides of an acute triangle with a circumradius of R.
Show that R^6 ≥ (4R² - a²)(4R² - b²)(4R² - c²).

Answer 36.
R^6 ≥ (4R² - a²)(4R² - b²)(4R² - c²)
<=> R^6≥(4R² - 4R^2sin^2 A)(4R² - 4R^2sin^2 B)(4R² - 4r^2 sin^2 C)
<=> 1 ≥ 64(1 - sin^2 A)(1 - sin^2 B)(1 - sin^2 C)
<=> 1/64 ≥ cos^2 A * cos^2 B * cos^2 C
<=> 1/8 ≥ lcosA cosB cos Cl
<=> 1 - 8 lcosA cosB cosCl ≥ 0
<=> 1 - 8cosAcosBcosC ≥ 0 [for acute triangle]
<=> 1 - 4(2cosAcosB)cosC ≥ 0
<=> 1 - 4[cos(A+B) + cos(A-B)]cosC ≥ 0
<=> 1 - 4[cos(π-C) + cos(A-B)]cosC ≥ 0
<=> 1 - 4[-cosC + cos(A-B)]cosC ≥ 0
<=> 4cos^2 C - 4cosCcos(A-B) + 1 ≥ 0
<=> [2cosC - cos(A-B)]^2 + 1 - cos^2 (A-B) ≥ 0
<=> [2cosC - cos(A-B)]^2 + sin^2 (A-B) ≥ 0
which is true
=> original inequality is true for acute triangle.

LINK to YA!