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Question-152365




Question Number 152365 by mathdanisur last updated on 27/Aug/21
Answered by ghimisi last updated on 28/Aug/21
a=x+y;b=y+z;c=x+z  p=x+y+z;q=ab+bc+ac;r=abc  ⇔....⇔p^3 +9r≥4pq⇔schur
$${a}={x}+{y};{b}={y}+{z};{c}={x}+{z} \\ $$$${p}={x}+{y}+{z};{q}={ab}+{bc}+{ac};{r}={abc} \\ $$$$\Leftrightarrow….\Leftrightarrow{p}^{\mathrm{3}} +\mathrm{9}{r}\geqslant\mathrm{4}{pq}\Leftrightarrow{schur} \\ $$
Commented by mathdanisur last updated on 28/Aug/21
Thank you Ser nice
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser}\:\mathrm{nice} \\ $$

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