CAT — Inequalities

Step 1: Express both equations in simpler form
Let x² + y² + z² = S. Then a = 4S.
Also, x − y − z = (3 + a)/4 = (3 + 4S)/4 = S + 3/4

Step 2: Apply Cauchy–Schwarz Inequality
(x − y − z)² ≤ (1² + (−1)² + (−1)²)(x² + y² + z²)
(x − y − z)² ≤ 3S
Substituting x − y − z = S + 3/4:
(S + 3/4)² ≤ 3S

Step 3: Simplify
S² + (3/2)S + 9/16 ≤ 3S
S² − (3/2)S + 9/16 ≤ 0
16S² − 24S + 9 ≤ 0
(4S − 3)² ≤ 0
Since a square is always non-negative, this is possible only when 4S − 3 = 0 → S = 3/4

Step 4: Find a
a = 4S = 4 × (3/4) = 3

Step 1: Bring both fractions to one side
x/y − (x + 3)/(y − 3) < 0
Taking LCM: [x(y − 3) − y(x + 3)] / [y(y − 3)] < 0
Simplifying the numerator: xy − 3x − xy − 3y = −3(x + y)
Hence: −3(x + y) / [y(y − 3)] < 0
Dividing by −3 reverses the inequality: (x + y) / [y(y − 3)] > 0
So the numerator and denominator must have the same sign.

Step 2: Check Option B
Given: y < 0 and −x < y → x + y > 0
Since y < 0 and (y − 3) < 0: y(y − 3) > 0
Numerator > 0 and Denominator > 0
Hence (x + y) / [y(y − 3)] > 0 ✓

Step 3: Check the other options
Option A: Does not guarantee the required sign condition.
Option C: y > 10 makes denominator positive, but −x > y implies x + y < 0, so inequality fails.
Option D: Does not always ensure numerator and denominator have the same sign.
Therefore, only Option B always satisfies the given inequality.

Step 1: Convert all logarithms to base 2. Since 64 = 2⁶, 8 = 2³, 512 = 2⁹: log₆₄ x² = (1/6) log₂ x² = (1/3) log₂ x. log₈ √y = (1/3) log₂ √y = (1/6) log₂ y. 3 log₅₁₂ (√yz) = 3 × (1/9) log₂ (√yz) = (1/3) × (1/2) log₂ (yz) = (1/6)(log₂ y + log₂ z). Substituting: (1/3) log₂ x + (1/6) log₂ y + (1/6) log₂ y + (1/6) log₂ z = 4. So, (1/3) log₂ x + (1/3) log₂ y + (1/6) log₂ z = 4. Multiply throughout by 6: 2 log₂ x + 2 log₂ y + log₂ z = 24. Using logarithm properties: log₂ (x²y²z) = 24. Hence, x²y²z = 2²⁴.

Step 2: Apply AM-GM Inequality. Apply AM-GM directly to x², y² and z: (x² + y² + z)/3 ≥ (x²y²z)^(1/3) = 2⁸ = 256. Equality holds when x² = y² = z = 256. Since x and y are positive: x = 16, y = 16, z = 16. Therefore, x + y + z = 16 + 16 + 16 = 48.

Given:
1/(x + 5) ≤ 1/(2x − 3)
Find all values of x.

Step 1: Note the restrictions
The denominators cannot be zero.
So,
x ≠ −5
and
x ≠ 3/2

Step 2: Bring all terms to one side
1/(x + 5) − 1/(2x − 3) ≤ 0
Taking the LCM,
[(2x − 3) − (x + 5)] / [(x + 5)(2x − 3)] ≤ 0
(x − 8) / [(x + 5)(2x − 3)] ≤ 0

Step 3: Find the critical points
The critical points are
x = −5, 3/2, 8
These divide the number line into four intervals:
● x < −5
● −5 < x < 3/2
● 3/2 < x < 8
● x > 8

Step 4: Check the sign in each interval
For x < −5
Choose x = −6.
Expression = (−)/(+)
= Negative
Satisfied.
For −5 < x < 3/2
Choose x = 0.
Expression = (−)/(−)
= Positive
Not satisfied.
For 3/2 < x < 8
Choose x = 2.
Expression = (−)/(+)
= Negative
Satisfied.
For x > 8
Choose x = 9.
Expression = (+)/(+)
= Positive
Not satisfied.

Step 5: Check the boundary point
At x = 8,
the numerator becomes zero while the denominator is non-zero.
Hence,
the inequality is satisfied.
The points
x = −5
and
x = 3/2
are not included since the expression is undefined.

Final Answer
The solution set is
x < −5
or
3/2 < x ≤ 8

Answer:
A. x < −5 or 3/2 < x ≤ 8

Step 1: p + q + r = 900, 150 < r < 500, r is a perfect square.
Let S = 900 − r, so p + q = S.

Step 2: Constraints on p
p ≥ 0.3q = 0.3(S−p) → 1.3p ≥ 0.3S → p ≥ 3S/13
p ≤ 0.7q = 0.7(S−p) → 1.7p ≤ 0.7S → p ≤ 7S/17

Step 3: Maximize p → maximize S → minimize r.
Smallest perfect square > 150: r = 169 → S = 731
Maximum p = ⌊7×731/17⌋ = ⌊5117/17⌋ = 301

Step 4: Minimize p → minimize S → maximize r.
Largest perfect square < 500: r = 484 → S = 416
Minimum p = ⌈3×416/13⌉ = ⌈1248/13⌉ = 96

Step 5: Sum = 301 + 96 = 397