Prove that 12(ab+ba+ac)&lt;7a2+15b2+18c2 holds for all positive numbers.

Step 1By AM-GM, or completing the square,(p−q)2≥0⇒p2+q2≥2pqwe have(4a2+9b2)+(3a2+12c2)+(6b2+6c2)≥12ab+12ca+12bcwith equality for 2a=3b, a=2c, b=c whose simultaneous solution is (a,b,c)=(0,0,0)For a,b,c&gt;0, we have strict inequality.

Step 1We writef(a,b,c)=7a2+15b2+18c2−12(ab+bc+ca)as a quadratic form with matrix[7−6−6−615−6−6−618]This can be verified to be a positive-definite matrix, so f(a,b,c)≥0 for all a,b,c∈R. In particular, if a,b,c&gt;0 then f(a,b,c)&gt;0.

Step 1By AM-GM, or completing the square,(p−q)2≥0⇒p2+q2≥2pqwe have(4a2+9b2)+(3a2+12c2)+(6b2+6c2)≥12ab+12ca+12bcwith equality for 2a=3b, a=2c, b=c whose simultaneous solution is (a,b,c)=(0,0,0)For a,b,c&gt;0, we have strict inequality.

Step 1We writef(a,b,c)=7a2+15b2+18c2−12(ab+bc+ca)as a quadratic form with matrix[7−6−6−615−6−6−618]This can be verified to be a positive-definite matrix, so f(a,b,c)≥0 for all a,b,c∈R. In particular, if a,b,c&gt;0 then f(a,b,c)&gt;0.

Explained: "Prove that 12(ab+ba+ac)<7a2+15b2+18c2 holds for all positive numbers."

High School
Algebra
Algebra I
Quadratic function and equation

David Rodgers

Answered question

2022-03-28

Prove that

12 (a b + b a + a c) < 7 a^{2} + 15 b^{2} + 18 c^{2}

holds for all positive numbers.

Answer & Explanation

kattylouxlvc

Beginner2022-03-29Added 11 answers

Step 1
By AM-GM, or completing the square,

{(p - q)}^{2} \geq 0 \Rightarrow p^{2} + q^{2} \geq 2 p q

we have

(4 a^{2} + 9 b^{2}) + (3 a^{2} + 12 c^{2}) + (6 b^{2} + 6 c^{2}) \geq 12 a b + 12 c a + 12 b c

with equality for

2 a = 3 b, a = 2 c, b = c

whose simultaneous solution is

(a, b, c) = (0, 0, 0)

For

a, b, c > 0

, we have strict inequality.

Raiden Griffin

Beginner2022-03-30Added 13 answers

Step 1
We write

f (a, b, c) = 7 a^{2} + 15 b^{2} + 18 c^{2} - 12 (a b + b c + c a)

as a quadratic form with matrix

[\begin{matrix} 7 & - 6 & - 6 \\ - 6 & 15 & - 6 \\ - 6 & - 6 & 18 \end{matrix}]

This can be verified to be a positive-definite matrix, so

f (a, b, c) \geq 0

for all

a, b, c \in R

. In particular, if

a, b, c > 0

then

f (a, b, c) > 0

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