Question Number 204187 by mnjuly1970 last updated on 08/Feb/24
$$ \\ $$$$\:\:\:\:\:\:{If}\:\:\:{p}\:,\:{q}\:,\:{r}\:>\mathrm{0}\:,\:\:{pqr}=\:\mathrm{1} \\ $$$$\:\:\:\:\: \\ $$$$\:\Rightarrow\:\:{min}\left(\frac{\:{p}^{\mathrm{3}} }{{q}+{r}}\:+\:\frac{{q}^{\:\mathrm{3}} }{{p}+{r}}\:+\frac{{r}^{\mathrm{3}} }{{p}+{q}}\:\right)=? \\ $$$$\:\:\:\ast\:{give}\:{a}\:{reason}\:\ast \\ $$
Answered by deleteduser1 last updated on 08/Feb/24
![Σ(p^3 /(q+r))=Σ(p^4 /(pq+pr))≥(((p^2 +q^2 +r^2 )^2 )/(2(pq+qr+rp)))≥(((p^2 +q^2 +r^2 )^2 )/(2(p^2 +q^2 +r^2 ))) =((p^2 +q^2 +r^2 )/2)≥((3(((pqr)^2 ))^(1/3) )/2)=(3/2)[Equality when p=q=r=1]](Q204190.png)
$$\Sigma\frac{{p}^{\mathrm{3}} }{{q}+{r}}=\Sigma\frac{{p}^{\mathrm{4}} }{{pq}+{pr}}\geqslant\frac{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({pq}+{qr}+{rp}\right)}\geqslant\frac{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} \right)} \\ $$$$=\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} }{\mathrm{2}}\geqslant\frac{\mathrm{3}\sqrt[{\mathrm{3}}]{\left({pqr}\right)^{\mathrm{2}} }}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\left[{Equality}\:{when}\:{p}={q}={r}=\mathrm{1}\right] \\ $$
Commented by mnjuly1970 last updated on 08/Feb/24
$$\:{T} \\ $$
Answered by witcher3 last updated on 08/Feb/24
$$\mathrm{p}>\mathrm{q}>\mathrm{r} \\ $$$$\frac{\mathrm{1}}{\mathrm{q}+\mathrm{r}}>\frac{\mathrm{1}}{\mathrm{p}+\mathrm{r}}>\frac{\mathrm{1}}{\mathrm{p}+\mathrm{q}} \\ $$$$\mathrm{reardement}\:\mathrm{inequality} \\ $$$$\mathrm{LHs}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{q}^{\mathrm{3}} +\mathrm{r}^{\mathrm{3}} }{\mathrm{q}+\mathrm{r}}+\frac{\mathrm{p}^{\mathrm{3}} +\mathrm{r}^{\mathrm{3}} }{\mathrm{p}+\mathrm{r}}+\frac{\mathrm{p}^{\mathrm{3}} +\mathrm{q}^{\mathrm{3}} }{\mathrm{p}+\mathrm{q}}\right) \\ $$$$\mathrm{LHS}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2q}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} +\mathrm{2p}^{\mathrm{2}} −\mathrm{qr}−\mathrm{pr}−\mathrm{pq}\right) \\ $$$$\left(\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{p}^{\mathrm{2}} \right)\underset{\mathrm{CS}} {\geqslant}\mathrm{pr}+\mathrm{rq}+\mathrm{qp} \\ $$$$\mathrm{LHS}\geqslant\frac{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\geqslant\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{p}^{\mathrm{2}} \mathrm{q}^{\mathrm{2}} \mathrm{r}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Commented by mnjuly1970 last updated on 08/Feb/24
$${thanks}\:{alot}\:{sir}\:\:\lesseqgtr \\ $$
Commented by witcher3 last updated on 08/Feb/24
$$\mathrm{withe}\:\mathrm{Pleasur} \\ $$