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Sa se arate ca daca a,b,c,d>0 si

\( \array\begin\{a\le1\\a+b\le 5\\a+b+c\le14\\a+b+c+d\le30 \)

atunci \( \sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10 \)

Imi cer scuze daca n-o sa fie bine dar e o rezolvare la nivelul clasei a VI-a
Din ce-am studiat inainte cu radicalii :
\( a\le 1 \Rightarrow \sqrt{a}\le 1 \)
\( a\le 1 \Rightarrow 1+b\le 5 \Rightarrow b\le 4 \Rightarrow \sqrt{b}\le 2 \)
\( a+b\le 5 \Rightarrow 5+c\le 14 \Rightarrow c\le 9 \Rightarrow \sqrt{c}\le 3 \)
\( a+b+c \le 14 \Rightarrow 14+d\le 30 \Rightarrow d\le 16 \Rightarrow \sqrt{d}\le 4 \)
Adunandu-le obtinem : \( \sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\le 10 \)
Nu prea stapanesc radicalii , dar sper sa fie bine .

A ... si chestia asta cu radicalii merge numai daca numerele sunt pozitive (\( a\le 4 \Rightarrow \sqrt{a}\le 2 \) doar daca \( a>0 \)) . Altfel nu merge , nu ?

Quit wrote:Imi cer scuze daca n-o sa fie bine dar e o rezolvare la nivelul clasei a VI-a
Din ce-am studiat inainte cu radicalii :
\( a\le 1 \Rightarrow \sqrt{a}\le 1 \)
\( a\le 1 \Rightarrow 1+b\le 5 \Rightarrow b\le 4 \Rightarrow \sqrt{b}\le 2 \)

,,Chestia '' asta nu prea merge,

ia de exemplu x=0,5 si b=5

\( x=a\leq1,y=a+b\leq5,z=a+b+c\leq14,t=a+b+c+d\leq30.\\
\sum\sqrt{a}=\sqrt{a}+\frac{\sqrt{4b}}{2}+\frac{\sqrt{9c}}{3}+\frac{\sqrt{16d}}{4}\leq\frac{1}{2}(a+1+\frac{b+4}{2}+\frac{c+9}{3}+\frac{d+16}{4})
=\frac{1}{2}(\frac{12a+6b+4c+3d}{12}+10)=\\=\frac{1}{2}(\frac{12x+6(y-x)+4(z-y)+3(t-z)}{12}+10)=
\frac{1}{2}(\frac{6x+2y+z+3t}{12}+10)\leq\frac{1}{2}(\frac{6\cdot1+2\cdot5+14+3\cdot30}{12}+10=\frac{1}{2}(\frac{120}{12}+10)=10 \).
Egalitate cand \( a=1,b=4,c=9,d=16 \).