If a,b,c are distinct real numbers,such that the quadratic expression Q1 (x)=ax^2+bx+c,Q2 (x)=bx^2+cx+a and Q3 (x)=cx^2+ax+b are always non-negative,then the possible integer in the range of the expression y=a^2+b^2+c^2/ab+bc+ac
(a) 1 (b) 2 (c) 3 (d) 4

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Point} \mathrm{to} \mathrm{remember}: \\ \mathrm{If} \mathrm{a} \mathrm{quadratic} \mathrm{equation} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{Px}}^2+\mathrm{Qx}+\mathrm{R} \mathrm{is} \mathrm{always} \mathrm{non}-\mathrm{negative}, \mathrm{then} \\ \mathrm{P}>0 \mathrm{and} \mathrm{Discriminat}={\mathrm{Q}}^2-4\mathrm{PR}\ge 0 \mathrm{or} {\mathrm{Q}}^2>4\mathrm{PR} \\ \mathrm{Now} \\ \mathrm{given} \mathrm{quadratic} \mathrm{equations} \mathrm{are} \mathrm{always} \mathrm{non}-\mathrm{negative}. \\ {\mathrm{Q}}_1\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \Rightarrow \mathrm{a}>0 \mathrm{and} {\mathrm{b}}^2\ge 4\mathrm{ac} ;\left(\mathrm{i}\right) \\ {\mathrm{Q}}_2\left(\mathrm{x}\right)={\mathrm{bx}}^2+\mathrm{cx}+\mathrm{a} \\ \Rightarrow \mathrm{b}>0 \mathrm{and} {\mathrm{c}}^2\ge 4\mathrm{ab} ;\left(\mathrm{ii}\right) \\ {\mathrm{Q}}_3\left(\mathrm{x}\right)={\mathrm{cx}}^2+\mathrm{ax}+\mathrm{b} \\ \Rightarrow \mathrm{c}>0 \mathrm{and} {\mathrm{a}}^2\ge 4\mathrm{bc} ;\left(\mathrm{iii}\right) \\ \mathrm{By} \left(\mathrm{i}\right), \left(\mathrm{ii}\right) \mathrm{and} \left(\mathrm{iii}\right), \mathrm{we} \mathrm{get}: \\ \mathrm{a}>0 \mathrm{and} \mathrm{b}>0 \mathrm{and} \mathrm{c}>0 \\ \mathrm{Now} \\ {\mathrm{a}}^2\ge 4\mathrm{bc} \\ {\mathrm{b}}^2\ge 4\mathrm{ac} \\ {\mathrm{c}}^2\ge 4\mathrm{ab} \\ \mathrm{Adding} \mathrm{we} \mathrm{get} \\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\ge 4\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right) \\ \mathrm{As} \mathrm{a}>0 \mathrm{and} \mathrm{b}>0 \mathrm{and} \mathrm{c}>0, \mathrm{hence} \mathrm{ab}+\mathrm{bc}+\mathrm{ca}>0 \\ \mathrm{Hence} \\ {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\ge 4\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right) \mathrm{becomes}: \\ \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}\ge 4 \\ \mathrm{Hence} \mathrm{one} \mathrm{of} \mathrm{the} \mathrm{possible} \mathrm{values} \mathrm{of} \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}} \mathrm{is} 4. \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{d}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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