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I do not think that you can add the inequalities. The thing is that the three inequalities that you have added do not take minimum value for the same x and y. If all the three took minimum value for same x and y, then we could say that LHS is greater than or equal to 17root2. But now we can only say that it is strictly greater than 17root2.

Indeed. What he has shown is that $17 \sqrt{2}$ is a lower bound. However, he has not shown that this is the maximum possible lower bound, which would then be the minimum of the function.

For example, it is obvious that $\sqrt{ x^2 + 144 } \geq 0$ and $\sqrt{ y^2 + 25 } \geq 0$, which tells us that $\sqrt{ x^2 + 144 } + \sqrt{ y^2 + 25 } \geq 0$. But clearly, 0 is merely a lower bound of the expression, and is not equal to the minimum value of this function.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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TopNewestI do not think that you can add the inequalities. The thing is that the three inequalities that you have added do not take minimum value for the same x and y. If all the three took minimum value for same x and y, then we could say that LHS is greater than or equal to 17root2. But now we can only say that it is strictly greater than 17root2.

Thank you for taking the time to put this up.

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Indeed. What he has shown is that $17 \sqrt{2}$ is a lower bound. However, he has not shown that this is the maximum possible lower bound, which would then be the minimum of the function.

For example, it is obvious that $\sqrt{ x^2 + 144 } \geq 0$ and $\sqrt{ y^2 + 25 } \geq 0$, which tells us that $\sqrt{ x^2 + 144 } + \sqrt{ y^2 + 25 } \geq 0$. But clearly, 0 is merely a lower bound of the expression, and is not equal to the minimum value of this function.

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Ok I've understood this. And trying the problem to solve geometrically.

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