@Lia - Maths is fun !\(Let:a,b,c\ge0\text{ }such:a+b+c=3.Found\text{ }max\text{ }and\text{ }min\text{ }A=\sqrt{x+3}+\sqrt{y+3}+\sqrt{z+3}\) My solution !*Found maxUsing Bunhiacopxki we have\(A^2\le\left(a+3+b+3+c+3\right)\left(1+1+1\right)=...=36\)\(\Rightarrow A\le6\left(Because\:\text{ }\text{ }A\ge0\text{ }so\text{ }A\text{ }can't\text{ }< 0\text{ }\right)\)\(A_{max}=6\text{ }\Leftrightarrow a=b=c=1\)*Found minWe have extra...
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@Lia - Maths is fun !
\(Let:a,b,c\ge0\text{ }such:a+b+c=3.Found\text{ }max\text{ }and\text{ }min\text{ }A=\sqrt{x+3}+\sqrt{y+3}+\sqrt{z+3}\)
My solution !
*Found max
Using Bunhiacopxki we have
\(A^2\le\left(a+3+b+3+c+3\right)\left(1+1+1\right)=...=36\)
\(\Rightarrow A\le6\left(Because\:\text{ }\text{ }A\ge0\text{ }so\text{ }A\text{ }can't\text{ }< 0\text{ }\right)\)
\(A_{max}=6\text{ }\Leftrightarrow a=b=c=1\)
*Found min
We have extra inequality \(\sqrt{x+z}+\sqrt{y+z}\ge\sqrt{z}+\sqrt{x+y+z}\left(x;y;z\ge0\right)\)(1)
Prove : \(\left(1\right)\Leftrightarrow x+y+2z+2\sqrt{\left(x+z\right)\left(y+z\right)}\ge z+x+y+z+2\sqrt{z\left(x+y+z\right)}\)
\(\Leftrightarrow\sqrt{xy+xz+yz+z^2}\ge\sqrt{xz+yz+z^2}\)
\(\Leftrightarrow xy+xz+yz+z^2\ge xz+yz+z^2\)
\(\Leftrightarrow xy\ge0\left(True!\right)\)
Using (1) we have
\(A=\sqrt{a+3}+\sqrt{b+3}+\sqrt{c+3}\ge\sqrt{3}+\sqrt{a+b+3}+\sqrt{c+3}\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{a+b+c}\)
\(=3\sqrt{3}\)
\(A_{min}=3\sqrt{3}\text{ }when\text{ }\hept{\begin{cases}a=b=\frac{3}{2}\\c=0\end{cases}}\)
(In here I using when because there are many other a,b,c such a = 0 ; b = c = 3/2)
The problem is done !
