Question

Daca x,y>0, a,b ∈ R, atunci $\frac{a^{2}}{x} +\frac{b^{2}}{y} \geq \frac{(a+b)^2}{x+y}$. Deduceti $\frac{a^{2}}{x} +\frac{b^{2}}{y} +\frac{c^{2}}{z} \geq \frac{(a+b+c)^2}{x+y+z}$

Answer 1

[tex]\text{Vom demonstra inegalitatea prin calcul direct.}\\ \dfrac{a^2}{x}+\dfrac{b^2}{y}\geq \dfrac{(a+b)^2}{x+y}|\cdot (x+y)\\ \dfrac{a^2(x+y)}{x}+\dfrac{b^2(x+y)}{y} \geq (a+b)^2\\ \dfrac{a^2\cdot x+a^2\cdot y}{x}+\dfrac{b^2\cdot x+b^2\cdot y}{y}\geq a^2+2ab+b^2\\ a^2+\dfrac{a^2\cdot y}{x}+b^2+ \dfrac{b^2\cdot x}{y}\geq a^2+b^2+2ab\\ a^2\cdot \dfrac{y}{x}+b^2\cdot \dfrac{x}{y}\geq 2ab [/tex][tex]\text{Din inegalitatea mediilor avem ca:}\\ a^2\cdot \dfrac{y}{x}+b^2\cdot \dfrac{x}{y}\geq 2\sqrt{a^2\cdot \dfrac{y}{x}\cdot b^2\cdot \dfrac{x}{y}}= 2ab ,c.c.t.d.\\ \text{Mai departe avem ca :}\\ \dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\geq \dfrac{(a+b)^2}{x+y}+\dfrac{c^2}{z}\geq \dfrac{(a+b+c)^2}{x+y+z}, Q.E.D[/tex]