UvU à nhầm u;v;w chứ @@
\(\left(x+y+z;xy+zx+yz;xyz\right)->\left(3u;3v^2;w^3\right)\)
ta can cm\(w\le\dfrac{u}{\sqrt[3]{2}}\) voi \(9u^2=12v^2\)
notethat: dieu kien da cho ko co \(w\) nen ta co the k,dinh rang co the tim dc gia tri lon nhat cua \(w^3\), xay ra khi 2 bien bang nhau. WLOg x=y
\(gt->z\left(z-4x\right)=0\)
+)z=0 bdt luon dung
+)z=4x ta cco bdt can cm \(5x+y\ge3\sqrt[3]{8x^2y}\)
\(\Leftrightarrow\left(5x+y\right)^3-\left(6\sqrt[3]{x^2y}\right)^3\ge0\)
\(\Leftrightarrow\left(x-y\right)\left(125x^2-16xy-y^2\right)\ge0\)
\(\Leftrightarrow0\ge0\)
True af
coi \(x^2+y^2+z^2=2xy+2yz+2xz\) la pt bac 2 an \(z\)
(delta,nhan chia cac thu....)
\(\left[{}\begin{matrix}z=x+y+2\sqrt{xy}\\z=x+y-2\sqrt{xy}\end{matrix}\right.\)
+)\(z=x+y-2\sqrt{xy}\). ta cần cm \(2\left(x+y-\sqrt{xy}\right)\ge3\sqrt[3]{2xy\left(x+y-2\sqrt{xy}\right)}\)
\(\left(\sqrt{x};\sqrt{y}\right)->\left(a;b\right)\) (cho gọn)
\(\left(2\left(a^2+b^2-ab\right)\right)^3-\left(3\sqrt[3]{2a^2b^2\left(a^2+b^2-2ab\right)}\right)^3\ge0\)
\(\Leftrightarrow2\left(a+b\right)^2\left(2a-b\right)^2\left(a-2b\right)^2\ge0\)
+)\(z=x+y+2\sqrt{xy}\) cũng cần cm
\(2\left(x+y+\sqrt{xy}\right)\ge3\sqrt[3]{2xy\left(x+y+2\sqrt{xy}\right)}\)
\(\left(\sqrt{x};\sqrt{y}\right)->\left(a;b\right)\)
\(\left(2\left(a^2+b^2+ab\right)\right)^3-\left(3\sqrt[3]{2a^2b^2\left(a^2+b^2+2ab\right)}\right)^3\ge0\)
\(\Leftrightarrow2\left(a-b\right)^2\left(2a+b\right)^2\left(a+2b\right)^2\ge0\)
Một lời giải nhẹ nhàng hơn dồn biến hay uvw:
\(x^2+y^2+z^2=2(xy+yz+xz)\Rightarrow (x+y+z)^2=4(xy+yz+xz)\)
\(\Rightarrow \frac{xy}{(x+y+z)^2}+\frac{yz}{(x+y+z)^2}+\frac{xz}{(x+y+z)^2}=\frac{1}{4}\)
Đặt \(\left(\frac{x}{x+y+z},\frac{y}{x+y+z}, \frac{z}{x+y+z}\right)=(a,b,c)\)
Khi đó ta có: \(\left\{\begin{matrix} a+b+c=1\\ ab+bc+ac=\frac{1}{4}\end{matrix}\right.\)
Điều ta cần phải chứng minh \(\Leftrightarrow (x+y+z)^3\geq 54xyz\Leftrightarrow \frac{xyz}{(x+y+z)^3}\leq \frac{1}{54}\Leftrightarrow abc\leq \frac{1}{54}\)
----------------------------------------------------------
Thật vậy:
Không mất tổng quát, giả sử \(a=\min (a,b,c)\Rightarrow 1=a+b+c\geq 3a\Rightarrow a\leq \frac{1}{3}\Rightarrow 3a-2<0\)
\(abc=a.bc=a(\frac{1}{4}-ab-ac)=a[\frac{1}{4}-a(b+c)]=a[\frac{1}{4}-a(1-a)]\)
\(=a^3-a^2+\frac{1}{4}a\)
Xét hiệu: \(abc-\frac{1}{54}=a^3-a^2+\frac{1}{4}a-\frac{1}{54}=\frac{108a^3-108a^2+27a-2}{108}=\frac{(3a-2)(6a-1)^2}{108}\leq 0\) do \(3a-2<0\)
\(\Rightarrow abc\leq \frac{1}{54}\)
Vậy ta có đpcm.
Dấu "=" xảy ra khi \((x,y,z)=(4n,n,n)\) với $n$ là một số không âm nào đó.
