If $\frac{a}{b}<\frac{a'}{b'}<\frac{a''}{b''}$, with all values positive, then $\frac{a}{b}<\frac{ab+a’b’+a’’b’’}{b^2+b’^2+b’’^2}<\frac{a’’}{b’’}$

Question

Bonjour. I need help for this.

Let $a$, $b$, $a^\prime$, $b^\prime$, $a^{\prime\prime}$, and $b^{\prime\prime}$ be six non-zero positive numbers such that $$\frac{a}{b}<\frac{a^\prime}{b^\prime}<\frac{a^{\prime\prime}}{b^{\prime\prime}}$$ We want to prove that this stuff holds: $$\frac{a}{b}<\frac{ab+a^\prime b^\prime+a^{\prime\prime}b^{\prime\prime}}{b^2+{b^{\prime}}^2+{b^{\prime\prime}}^2}<\frac{a^{\prime\prime}}{b^{\prime\prime}}$$

Thanks.

Answer 1

Let $\frac{a}{b}=k,$ $\frac{a'}{b'}=m$ and $\frac{a''}{b''}=n$,$b=x$, $b'=y$ and $b''=z$.

Thus, $k<m<n$ and we need to prove that $$k<\frac{kx^2+my^2+nz^2}{x^2+y^2+z^2}<n,$$ which is true because the left inequality it's $$(m-k)y^2+(n-k)z^2>0$$ and the right inequality it's $$(n-m)y^2+(n-k)x^2>0.$$

Answer 2

We multiply the left inequality by $$b^2+b_1^2+b_2^2>0$$ then we get $$a_1b_1b+a_2b_2b>ab_1^2+ab_2^2$$ and this is $$b_1(a_1b-ab_1)+b_2(a_2b-ab_2)>0$$ this is true since we have $$\frac{a_1}{b_1}>\frac{a}{b}$$ and $$\frac{a_2}{b_2}>\frac{a}{b}$$