Inegalitate in patru variabile

[alex2008]

Fie \( a,b,c>0 \) si \( k\ge 8 \). Sa se demonstreze ca

\( \frac{a}{\sqrt{a^2+kbc}}+\frac{b}{\sqrt{b^2+kca}}+\frac{c}{\sqrt{c^2+kab}}\ge \frac{3}{\sqrt{k+1}}. \)

[Marius Mainea]

Notand \( f(k)=\sum {\frac{a}{\sqrt{a^2+kbc}}}-\frac{3}{\sqrt{k+1}} \) se observa ca k=8 este punct de minim pentru f, adica \( f(k)\ge f(8 ) \).
Asadar este suficient sa demonstram inegalitatea data in 2001 la OIM:

\( \sum {\frac{a}{\sqrt{a^2+8bc}}}\ge 1 \).

[Claudiu Mindrila]

Asadar este suficient sa demonstram inegalitatea data in 2001 la OIM:

\( \sum {\frac{a}{\sqrt{a^2+8bc}}}\ge 1 \).

Conform inegalitatii Holder avem \( \left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)\left(\sum a\left(a^{2}+8bc\right)\right)\ge\left(a+b+c\right)^{3} \), adica \( \left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)^{2}\ge\frac{\left(a+b+c\right)^{3}}{a^{3}+b^{3}+c^{3}+24abc} \).

Problema revine acum la \( \sum\frac{a}{\sqrt{a^{2}+8bc}}\ge1\Longleftrightarrow\left(\sum\frac{a}{\sqrt{a^{2}+8bc}}\right)^{2}\ge1\Longleftrightarrow\frac{\left(a+b+c\right)^{3}}{a^{3}+b^{3}+c^{3}+24abc}\ge1\Longleftrightarrow\sum a^{3}+6abc+3\sum ab\left(a+b\right)\ge\sum a^{3}+24abc, \) care este echivalenta cu binecunoscuta inegalitate \( \sum ab\left(a+b\right)\ge6abc \).

[alex2008]

Daca \( a,b,c>0 \) si \( k\ge 3 \) se poate demonstra ca :

\( \frac{a}{\sqrt{a^2+kbc}}+\frac{b}{\sqrt{b^2+kca}}+\frac{c}{\sqrt{c^2+kab}}< 2 \)

[alex2008]

Daca \( a,b,c>0 \) si \( k\ge 3 \) se poate demonstra ca :

\( \frac{a}{\sqrt{a^2+kbc}}+\frac{b}{\sqrt{b^2+kca}}+\frac{c}{\sqrt{c^2+kab}}< 2 \)

Sa observam ca \( \sum_{cyc}\frac{a}{\sqrt{a^2+kbc}}\le \sum_{cyc}\frac{a}{\sqrt{a^2+3bc}} \)

Deci ar fi de ajuns sa demonstram inegalitatea pentru cazul \( k=3 \) . Fiind simetrica si omogena sa presupunem ca \( a\ge b\ge c \) si \( abc=1 \) . Avem

\( \sum_{cyc}\frac{a}{\sqrt{a^2+3bc}}=\sum_{cyc}\sqrt{\frac{a^3}{a^3+3}} \) .

Fie \( x=a^3 \ ,\ y=b^3 \ ,\ z=c^3 \Rightarrow xyz=1 \) si \( x\ge y\ge z \) .

Avem ca \( y+z\le 2x =\frac{2}{yz}\ , \ yz\le 1 \) si \( \sqrt{\frac{x}{x+3}}< 1 \) .

Deci ar fi de ajuns sa demonstram ca : \( \sqrt{\frac{y}{y+3}}+\sqrt{\frac{z}{z+3}}\le 1\Leftrightarrow \frac{y}{y+3}+\frac{z}{z+3}+2\sqrt{\frac{yz}{(y+3)(z+3)}}\le 1\Leftrightarrow 2\sqrt{\frac{yz}{(y+3)(z+3)}}\le \frac{9-yz}{(x+3)(z+3)}\Leftrightarrow 4yz(y+3)(z+3)\le (9-yz)^2 \) .

\(
4yz(y+3)(z+3)=4yz(yz+3(y+z)+9)\le 4yz(yz+\frac{6}{yz}+9)=4(yz)^2+24+36yz \le 64 \)

In acelasi timp \( yz\le 1\Leftrightarrow 9-yz\ge 8 \Leftrightarrow (9-yz)^2\ge 64 \) , deci inegalitatea e demonstrata .