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Question Number 204187 by mnjuly1970 last updated on 08/Feb/24

$$$$$$Ifp,q,r>0,pqr=1$$$$$$$$\Rightarrow min(\frac{p^3}{q+r}+\frac{q^3}{p+r}+\frac{r^3}{p+q})=?$$$$\ast giveareason\ast$$

Answered by deleteduser1 last updated on 08/Feb/24

$$\mathrm{\Sigma }\frac{p^3}{q+r}=\mathrm{\Sigma }\frac{p^4}{pq+pr}⩾\frac{{(p^2+q^2+r^2)}^2}{2(pq+qr+rp)}⩾\frac{{(p^2+q^2+r^2)}^2}{2(p^2+q^2+r^2)}$$$$=\frac{p^2+q^2+r^2}{2}⩾\frac{3\sqrt[3]{{\left(pqr\right)}^2}}{2}=\frac{3}{2}[Equalitywhenp=q=r=1]$$

Commented by mnjuly1970 last updated on 08/Feb/24

$$T$$

Answered by witcher3 last updated on 08/Feb/24

$$\mathrm{p}>\mathrm{q}>\mathrm{r}$$$$\frac{1}{\mathrm{q}+\mathrm{r}}>\frac{1}{\mathrm{p}+\mathrm{r}}>\frac{1}{\mathrm{p}+\mathrm{q}}$$$$\mathrm{reardement}\mathrm{inequality}$$$$\mathrm{LHs}⩾\frac{1}{2}(\frac{{\mathrm{q}}^3+{\mathrm{r}}^3}{\mathrm{q}+\mathrm{r}}+\frac{{\mathrm{p}}^3+{\mathrm{r}}^3}{\mathrm{p}+\mathrm{r}}+\frac{{\mathrm{p}}^3+{\mathrm{q}}^3}{\mathrm{p}+\mathrm{q}})$$$$\mathrm{LHS}⩾\frac{1}{2}({2\mathrm{q}}^2+{2\mathrm{r}}^2+{2\mathrm{p}}^2-\mathrm{qr}-\mathrm{pr}-\mathrm{pq})$$$$({\mathrm{q}}^2+{\mathrm{r}}^2+{\mathrm{p}}^2)\underset{\mathrm{CS}}{⩾}\mathrm{pr}+\mathrm{rq}+\mathrm{qp}$$$$\mathrm{LHS}⩾\frac{{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2}{2}⩾\frac{3}{2}{\left({\mathrm{p}}^2{\mathrm{q}}^2{\mathrm{r}}^2\right)}^{\frac{1}{3}}=\frac{3}{2}$$

Commented by mnjuly1970 last updated on 08/Feb/24

$$thanksalotsir⋚$$

Commented by witcher3 last updated on 08/Feb/24

$$\mathrm{withe}\mathrm{Pleasur}$$

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