In b grams of sugar water containing a gram of sugar, add m grams of sugar and the water will become sweet.

This fact gives us the mathematical proposition:

If b>a>0, then a/b<(a+m)/(b+m), where m>0 and m is a real number

Solution:

Chemically, the sugar water becomes sweeter, meaning that the mass fraction of sugar dissolved in the water becomes larger

The mass fraction of sugar in the sugar water that dissolves in the water = a. The mass fraction of sugar dissolved in water = a/b*100%

Add m grams of sugar, that is, the amount of solute in the solution increases, after increasing the mass of sugar dissolved in water = a + m

Similarly, the mass of the solution increases, after increasing the mass of the solution becomes b + m

So the mass fraction of sugar dissolved in water increases after adding m grams of sugar to = (a + m )/(b+m)*100%

The mass of the solution is greater than the mass of the solute is greater than 0, b>a>0

This leads to the conclusion that a/b<(a+m)/(b+m) (note that m takes the range of m>0, m is a real number)

From the mathematical point of view, we have to carry out the calculations, as follows:

(a+m)/(b+m)-a/b

=[(a+m)b-(b+m)a]/(b+m)b

=m(b-a)/(b+m)b

Where (b+m)b is the denominator, and (b+m)b>0

m(b-a) is the numerator, and m(b-a)>0 from b>a>0

So m(b-a)/(b +m)b >0

There is a numerator, b+m)b >0

So m(b-a)/(b +m)b >0

i.e. (a+m)/(b+m)-a/b>0

It can be seen that (a+m)/(b+m)>a/b

This leads to the conclusion that a/b <(a+m)/(b+m) (note that m takes on a range of values m>0 and m is a real number)

Proof of mean value inequality

There are many ways of proving this. Mathematical induction (first or reverse induction), Lagrange multiplier method, Chin Sheng inequality method, sorting inequality method, Cauchy's inequality method, etc.

The following is a good understanding of the method

Chin Sheng inequality method

Chin Sheng inequality: the upper convex function f(x), x1,x2,.... .xn are any n points of the function f(x) in the interval (a,b),

then we have: f[(x1+x2+... +xn)/n]≥1/n*[f(x1)+f(x2)+... +f(xn)]

Let f(x)=lnx, f(x) is an upper convex increasing function

So, ln[(x1+x2+... +xn)/n] ≥ 1/n*[ln(x1)+ln(x2)+... +ln(xn)]=lnn times √(x1*x2*... *xn)

That is (x1+x2+... +xn)/n≥n times √(x1*x2*... *xn)