In orice triunghi \( 3(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2})-(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq 0 \).
(A. W. Walker, The Mathematics Magazine, vol. 43, nr. 4/1970, pag. 226)
O noua inegalitate
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Claudiu Mindrila
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O noua inegalitate
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maxim bogdan
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Notam \( a^2=x,b^2=y,c^2=z \).Deci \( x,y,z \) sunt laturile unui triunghi.WLOG \( x\ge y\ge z \)
Inegalitatea devine:
\( 3(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\ge 3+\displaystyle\sum_{cyc}\frac{x}{y}+\frac{y}{x} \)
Adica:\( \displaystyle\sum_{cyc}\frac{2x-y-z}{y}\ge 0 \)
Stim ca exista numerele \( m,n,p \) astfel incat \( x=m+n,y=m+p,z=n+p \).Deci \( m\ge n\ge p \).Inegalitatea este echivalenta cu:
\( \displaystyle\sum_{cyc}\frac{m}{m+p}+\displaystyle\sum_{cyc}\frac{m}{n+p}\ge 2\displaystyle\sum_{cyc}\frac{m}{m+n} \)
Aceasta inegalitate rezulta imediat din rearanjamente(suma\( \displaystyle\sum_{cyc}\frac{m}{m+n} \) este minima)
Inegalitatea devine:
\( 3(\frac{x}{y}+\frac{y}{z}+\frac{z}{x})\ge 3+\displaystyle\sum_{cyc}\frac{x}{y}+\frac{y}{x} \)
Adica:\( \displaystyle\sum_{cyc}\frac{2x-y-z}{y}\ge 0 \)
Stim ca exista numerele \( m,n,p \) astfel incat \( x=m+n,y=m+p,z=n+p \).Deci \( m\ge n\ge p \).Inegalitatea este echivalenta cu:
\( \displaystyle\sum_{cyc}\frac{m}{m+p}+\displaystyle\sum_{cyc}\frac{m}{n+p}\ge 2\displaystyle\sum_{cyc}\frac{m}{m+n} \)
Aceasta inegalitate rezulta imediat din rearanjamente(suma\( \displaystyle\sum_{cyc}\frac{m}{m+n} \) este minima)