MHB Inequality involving area under a curve

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The discussion centers on proving the inequality that for every x in the interval (0, 1), the integral of the function sqrt(1 + (cos y)^2) from 0 to 1 is greater than the expression sqrt(x^2 + (sin x)^2). Participants explore various approaches to demonstrate this inequality, including analyzing the behavior of the integral and the function involved. The integral is shown to represent a specific area under the curve, while the right side involves trigonometric functions evaluated at x. The conversation emphasizes the need for rigorous mathematical proof to validate the inequality. Ultimately, the inequality is established as a true statement for the specified range of x.
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Prove that for every $x\in (0,\,1)$ the following inequality holds:

$\displaystyle \int_0^1 \sqrt{1+(\cos y)^2} dy>\sqrt{x^2+(\sin x)^2}$
 
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Clearly $\displaystyle \int_0^1 \sqrt{1+(\cos y)^2} dy \ge \int_0^x \sqrt{1+(\cos y)^2} dy$ for each fixed $x\in (0,\,1)$. Observe that $\displaystyle \int_0^x \sqrt{1+(\cos y)^2} dy$ is the arc length of the function $f(y)=\sin y$ on the interval $[0,\,x]$ which is clearly strictly greater than the length of the straight line between the points $(0,\,0)$ and $(x,\, \sin x)$ which in turn is equal to $\sqrt{x^2+(\sin x)^2}$.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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