Johnson Kelley
05/01/2023 · Primary School

5. Prove that \( a^{2}-6 a+9+b^{2} \geqslant 0 \) for all real values of \( a \) and \( b \). 6. Prove that for all real values of \( x \) \( \begin{array}{lll}\text { (i) } x^{2}+6 x+9 \geqslant 0 & \text { (ii) } x^{2}-10 x+25 \geqslant 0 & \text { (iii) } x^{2}+4 x+6>0\end{array} \)

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To prove the inequalities, we will analyze each expression step by step. ### Problem 5: Prove that \( a^{2}-6a+9+b^{2} \geqslant 0 \) for all real values of \( a \) and \( b \). We can rewrite the expression \( a^{2}-6a+9 \) as follows: \[ a^{2}-6a+9 = (a-3)^{2} \] This is a perfect square, and since the square of any real number is non-negative, we have: \[ (a-3)^{2} \geq 0 \] Now, since \( b^{2} \) is also a perfect square, we know: \[ b^{2} \geq 0 \] Combining these two results, we get: \[ (a-3)^{2} + b^{2} \geq 0 + 0 = 0 \] Thus, we conclude that: \[ a^{2}-6a+9+b^{2} \geq 0 \] for all real values of \( a \) and \( b \). ### Problem 6: Prove the following inequalities for all real values of \( x \): #### (i) \( x^{2}+6x+9 \geqslant 0 \) We can rewrite this expression as: \[ x^{2}+6x+9 = (x+3)^{2} \] Again, since this is a perfect square, we have: \[ (x+3)^{2} \geq 0 \] This holds for all real values of \( x \). Therefore: \[ x^{2}+6x+9 \geq 0 \] #### (ii) \( x^{2}-10x+25 \geqslant 0 \) We can rewrite this expression as: \[ x^{2}-10x+25 = (x-5)^{2} \] Similar to the previous case, this is also a perfect square, so we have: \[ (x-5)^{2} \geq 0 \] This holds for all real values of \( x \). Therefore: \[ x^{2}-10x+25 \geq 0 \] #### (iii) \( x^{2}+4x+6 > 0 \) To analyze this expression, we can complete the square: \[ x^{2}+4x+6 = (x+2)^{2} + 2 \] Here, \( (x+2)^{2} \) is a perfect square and is always non-negative, so: \[ (x+2)^{2} \geq 0 \] Thus, we have: \[ (x+2)^{2} + 2 > 0 \] since \( (x+2)^{2} \) is non-negative and we are adding 2 to it. Therefore, the expression \( x^{2}+4x+6 \) is strictly greater than 0 for all real values of \( x \). ### Conclusion We have proven all the required inequalities: 1. \( a^{2}-6a+9+b^{2} \geq 0 \) for all \( a, b \in \mathbb{R} \). 2. \( x^{2}+6x+9 \geq 0 \) for all \( x \in \mathbb{R} \). 3. \( x^{2}-10x+25 \geq 0 \) for all \( x \in \mathbb{R} \). 4. \( x^{2}+4x+6 > 0 \) for all \( x \in \mathbb{R} \).

Quick Answer

1. \( a^{2}-6a+9+b^{2} \geq 0 \) for all \( a, b \in \mathbb{R} \). 2. \( x^{2}+6x+9 \geq 0 \) for all \( x \in \mathbb{R} \). 3. \( x^{2}-10x+25 \geq 0 \) for all \( x \in \mathbb{R} \). 4. \( x^{2}+4x+6 > 0 \) for all \( x \in \mathbb{R} \).
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