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Áp dụng bất đẳng thức Bunyakovsky:
\(NL^2=\left(\sqrt{4x+2\sqrt{x}+1}+\sqrt{4y+2\sqrt{y}+1}+\sqrt{4z+2\sqrt{z}+1}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(4x+2\sqrt{x}+1+4y+2\sqrt{y}+1+4z+2\sqrt{z}+1\right)\)
\(=3\left(4x+4y+4z\right)+3\left(2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\right)+3\left(1+1+1\right)\)
\(=12\left(x+y+z\right)+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+9\)
\(=153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
Mặt khác,theo Bunyakovsky: \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)=36\)
\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le6\)
\(\Rightarrow153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le153+36=189\)
\(\Rightarrow NL\le\sqrt{189}\)
Dấu "=" xảy ra khi: \(x=y=z=4\)
ta có \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\Rightarrow x+y+z\le3\)
ta có :\(\sqrt{4x+5}=\frac{\sqrt{9\left(4x+5\right)}}{3}\le\frac{9+4x+5}{2\times3}=\frac{2x+7}{3}\)
tương tự ta sẽ có ; \(A\le\frac{2x+7}{3}+\frac{2y+7}{3}+\frac{2z+7}{3}=\frac{2}{3}\left(x+y+z\right)+7\le\frac{2}{3}\times3+7=9\)
Vậy GTLN của A=9
dấu bằng xảy ra khi x= y= z =1
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2\ge2\left(xy+yz+zx\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=3.3=9\)
\(\Rightarrow x+y+z\le3\).
\(A=\sqrt{4x+5}+\sqrt{4y+5}+\sqrt{4z+5}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(4x+5+4y+5+4z+5\right)}\)
\(=\sqrt{3\left[4\left(x+y+z\right)+15\right]}=9\)
Dấu \(=\)khi \(x=y=z=1\).
\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)
\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)
Tương tự và cộng lại:
\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)
\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)
\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)
\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)
\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)
\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)
\(=\sqrt{189}\)
Dấu "=" xảy ra <=> x = y = z = 4
Ta sẽ chứng minh
\(\sqrt{x^2+1}+2\sqrt{x}\le\frac{2+\sqrt{2}}{2}\left(x+1\right)\)
\(\Leftrightarrow\left(\sqrt{x^2+1}+2\sqrt{x}\right)^2\le\frac{3+2\sqrt{2}}{2}\left(x+1\right)^2\)
\(\Leftrightarrow\frac{1+2\sqrt{2}}{2}\left(x^2+1\right)-4\sqrt{x\left(x^2+1\right)}+\left(2\sqrt{2}-1\right)x\ge0\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-\sqrt{2x}\right)\left(\frac{1+2\sqrt{2}}{2}\sqrt{x^2+1}-\frac{4-\sqrt{2}}{2}\sqrt{x}\right)\ge0\)
BĐT trên luôn đúng do \(x^2+1\ge2x\)
Vậy ta có:\(\text{∑}\sqrt{x^2+1}+2\sqrt{x}\le\text{∑}\frac{2+\sqrt{2}}{2}\left(x+1\right)\le6+3\sqrt{2}\)
Đẳng thức xảy ra khi x=y=z=1
Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
Bài 1:
a) Bạn xem lại đề bài hộ mình.
b) Thực hiện biến đổi tương đương:
\((x+y+z)^2\leq 3(x^2+y^2+z^2)\)
\(\Leftrightarrow x^2+y^2+z^2+2(xy+yz+xz)\leq 3(x^2+y^2+z^2)\)
\(\Leftrightarrow 2(xy+yz+xz)\leq 2(x^2+y^2+z^2)\)
\(\Leftrightarrow 2(x^2+y^2+z^2)-2(xy+yz+xz)\geq 0\)
\(\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0\)
BĐT trên luôn đúng do \(\left\{\begin{matrix} (x-y)^2\geq 0\\ (y-z)^2\geq 0\\ (z-x)^2\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{R}\)
Do đó ta có đpcm.
Dấu bằng xảy ra khi \(x=y=z\)
Bài 2:
\(A=\sqrt{4x+2\sqrt{x}+1}+\sqrt{4y+2\sqrt{y}+1}+\sqrt{4z+2\sqrt{z}+1}\)
\(\Rightarrow 2A=\sqrt{16x+8\sqrt{x}+4}+\sqrt{16y+8\sqrt{y}+4}+\sqrt{16z+8\sqrt{z}+4}\)
\(=\sqrt{18x-2(\sqrt{x}-2)^2+12}+\sqrt{18y-2(\sqrt{y}-2)^2+12}+\sqrt{18z-2(\sqrt{z}-1)^2+12}\)
\(\Rightarrow 2A\leq \sqrt{18x+12}+\sqrt{18y+12}+\sqrt{18z+12}(1)\)
Áp dụng BĐT Bunhiacopxky:
\((\sqrt{18x+12}+\sqrt{18y+12}+\sqrt{18z+12})^2\leq [(18x+12)+(18y+12)+(18z+1)](1+1+1)\)
\(=3[18(x+y+z)+36]=756\)
\(\Rightarrow \sqrt{18x+12}+\sqrt{18y+12}+\sqrt{18z+12}\leq \sqrt{756}=6\sqrt{21}(2)\)
Từ \((1); (2)\Rightarrow 2A\leq 6\sqrt{21}\Rightarrow A\leq 3\sqrt{21}\)
Vậy \(A_{\max}=3\sqrt{21}\). Dấu bằng xảy ra khi \(x=y=z=4\)
\(A=\sum\sqrt{4x+2\sqrt{x}+1}\)
\(Max_A=+\infty\)
\("="x=y=z=+\infty\)