Solve a^2 b^2-c^2>=a*b - ScanMath

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Type your math problem... a^2 b^2-c^2>=a*b

Question

Solve the inequality

Solve for a

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Solve for b

Solve for c

a \geq \frac{b 1 + 4 c ^{2} + ∣ b ∣}{2 ∣ b ∣ \times b} a \leq \frac{- b 1 + 4 c ^{2} + ∣ b ∣}{2 ∣ b ∣ \times b}

Evaluate

a^{2} b^{2} - c^{2} \geq ab

Rewrite the expression

b^{2} a^{2} - c^{2} \geq ba

Move the expression to the left side

b^{2} a^{2} - c^{2} - ba \geq 0

Move the constant to the right side

b^{2} a^{2} - ba \geq 0 - (- c^{2})

Add the terms

b^{2} a^{2} - ba \geq c^{2}

Evaluate

a^{2} - \frac{1}{b} \times a \geq \frac{c ^{2}}{b ^{2}}

Add the same value to both sides

a^{2} - \frac{1}{b} \times a + \frac{1}{( 2 b ) ^{2}} \geq \frac{c ^{2}}{b ^{2}} + \frac{1}{( 2 b ) ^{2}}

Evaluate

a^{2} - \frac{1}{b} \times a + \frac{1}{( 2 b ) ^{2}} \geq \frac{1 + 4 c ^{2}}{( 2 b ) ^{2}}

Evaluate

(a - \frac{1}{2 b})^{2} \geq \frac{1 + 4 c ^{2}}{( 2 b ) ^{2}}

Take the 2 -th root on both sides of the inequality

(a - \frac{1}{2 b})^{2} \geq \frac{1 + 4 c ^{2}}{( 2 b ) ^{2}}

Calculate

a - \frac{1}{2 b} \geq \frac{1 + 4 c ^{2}}{2 ∣ b ∣}

Separate the inequality into 2 possible cases

a - \frac{1}{2 b} \geq \frac{1 + 4 c ^{2}}{2 ∣ b ∣} a - \frac{1}{2 b} \leq - \frac{1 + 4 c ^{2}}{2 ∣ b ∣}

Calculate

More Steps

a \geq \frac{b 1 + 4 c ^{2} + ∣ b ∣}{2 ∣ b ∣ \times b} a - \frac{1}{2 b} \leq - \frac{1 + 4 c ^{2}}{2 ∣ b ∣}

Solution

More Steps

a \geq \frac{b 1 + 4 c ^{2} + ∣ b ∣}{2 ∣ b ∣ \times b} a \leq \frac{- b 1 + 4 c ^{2} + ∣ b ∣}{2 ∣ b ∣ \times b}

Show Solution