Daca \( x_1,x_2,x_3\in[a,b] \) unde \( 0<a<b \) atunci
\( (x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})\le 9+2\(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)^2 \)
Produsul MA x MH este marginit
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Marius Mainea
- Gauss
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Mateescu Constantin
- Newton
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Presupunem, fara a restrange generalitatea ca \( x_1\le x_2\le x_3 \) .
Aratam ca : \( \overline{\underline{\left\|\ (x_1+x_2+x_3)\left\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+x_2+x_3)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{x_3}\right\)=A\ \right\|}} \) .
Inegalitatea de mai sus este echivalenta succesiv cu urmatoarele:
\( 1+x_1\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)+(x_2+x_3)\frac{1}{x_1}+(x_2+x_3)\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ 1+a\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)+(x_2+x_3)\frac 1a+(x_2+x_3)\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\) \)
\( \Longleftrightarrow\ \frac{x_1(x_2+x_3)}{x_2x_3}+\frac{x_2+x_3}{x_1}\ \le\ \frac{a(x_2+x_3)}{x_2x_3}+\frac{x_2+x_3}{a}\ \Longleftrightarrow\ \frac{x_1}{x_2x_3}+\frac{1}{x_1}\ \le\ \frac{a}{x_2x_3}+\frac 1a \)
\( \Longleftrightarrow\ \(x_1-a)(x_1a-x_2x_3)\le 0 \) , adevarata deoarece \( 0<a\le x_1\le x_2\le x_3\le b \) .
In continuare demonstram ca : \( \overline{\underline{\left\|\ A=(a+x_2+x_3)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+x_2+b)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{b}\right\)=B\ \right\|}} \) .
Efectuand calculele ca mai sus ea se reduce la : \( (a+x_2)(x_3-b)(x_3b-ax_2)\le 0 \) , care are loc .
Analog se verifica si inegalitatea : \( \overline{\underline{\left\|\ B=(a+x_2+b)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{b}\right\)\ \le\ (a+b+a)\left\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}\right\)\ \right\|}}\ \Longleftrightarrow\ (a+b)(x_2-a)(x_2-b)\le 0 \) .
Din cele trei inegalitati de mai sus obtinem ca \( (x_1+x_2+x_3)\left\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+b+a)\left\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}\right\)=9+2\(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)^2 \) .
Aratam ca : \( \overline{\underline{\left\|\ (x_1+x_2+x_3)\left\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+x_2+x_3)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{x_3}\right\)=A\ \right\|}} \) .
Inegalitatea de mai sus este echivalenta succesiv cu urmatoarele:
\( 1+x_1\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)+(x_2+x_3)\frac{1}{x_1}+(x_2+x_3)\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ 1+a\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\)+(x_2+x_3)\frac 1a+(x_2+x_3)\left\(\frac{1}{x_2}+\frac{1}{x_3}\right\) \)
\( \Longleftrightarrow\ \frac{x_1(x_2+x_3)}{x_2x_3}+\frac{x_2+x_3}{x_1}\ \le\ \frac{a(x_2+x_3)}{x_2x_3}+\frac{x_2+x_3}{a}\ \Longleftrightarrow\ \frac{x_1}{x_2x_3}+\frac{1}{x_1}\ \le\ \frac{a}{x_2x_3}+\frac 1a \)
\( \Longleftrightarrow\ \(x_1-a)(x_1a-x_2x_3)\le 0 \) , adevarata deoarece \( 0<a\le x_1\le x_2\le x_3\le b \) .
In continuare demonstram ca : \( \overline{\underline{\left\|\ A=(a+x_2+x_3)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+x_2+b)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{b}\right\)=B\ \right\|}} \) .
Efectuand calculele ca mai sus ea se reduce la : \( (a+x_2)(x_3-b)(x_3b-ax_2)\le 0 \) , care are loc .
Analog se verifica si inegalitatea : \( \overline{\underline{\left\|\ B=(a+x_2+b)\left\(\frac{1}{a}+\frac{1}{x_2}+\frac{1}{b}\right\)\ \le\ (a+b+a)\left\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}\right\)\ \right\|}}\ \Longleftrightarrow\ (a+b)(x_2-a)(x_2-b)\le 0 \) .
Din cele trei inegalitati de mai sus obtinem ca \( (x_1+x_2+x_3)\left\(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}\right\)\ \le\ (a+b+a)\left\(\frac{1}{a}+\frac{1}{b}+\frac{1}{a}\right\)=9+2\(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\)^2 \) .