cho \(x,y\in R\) thỏa mãn \(x+y\ge1\). Tìm min \(A=\left(x^2+2y^2\right)\left(y^2+2x^2\right)-x^2-y^2-2x^2y^2\)
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1) Bất đẳng thức cần chứng minh
\(\Leftrightarrow\) a2 + b2 + c2 + d2 + \(2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge\left(a+c\right)^2+\left(b+d\right)^2\)
\(\Leftrightarrow\) \(ac+bd\le\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\left(1\right)\)
Nếu : ac + bd < 0 : BĐT luôn đúng
Nếu : ac + bd \(\ge\) 0 : Thì (1) tương đương
( ac + bd )2 \(\le\) ( a2 + b2 )( c2 + d2 )
\(\Leftrightarrow\) \(\left(ac\right)^2+\left(bd\right)^2+2abcd\le\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\)
\(\Leftrightarrow\) \(\left(ad\right)^2+\left(bc\right)^2-2abcd\ge0\)
\(\Leftrightarrow\) \(\left(ad-bc\right)^2\ge0\) , luôn đúng , vậy bài toán được chứng minh
2) Chọn :\(\left\{{}\begin{matrix}a=2\cos x.\cos y\\c=2\sin x.\sin y\\b=d=\sin\left(x-y\right)\end{matrix}\right.\)
Từ câu 1) ta có :
\(\sqrt{4\cos^2x.\cos^2y+\sin^2\left(x-y\right)}+\sqrt{4\sin^2x.\sin^2y+\sin^2\left(x-y\right)}\)
\(\ge\sqrt{\left(2\cos x.\cos y+2\sin x.\sin y\right)^2+\left(2\sin\left(x-y\right)\right)^2}\)
\(\ge\sqrt{4\cos^2\left(x-y\right)+4\sin^2\left(x-y\right)}=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chắc đề là \(x+y+z=3\)
Ta có:
\(\left(2x+y+z\right)^2=\left(x+y+x+z\right)^2\ge4\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P\le\dfrac{x}{4\left(x+y\right)\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)\left(y+z\right)}\)
\(\Rightarrow P\le\dfrac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{4\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\dfrac{xy+yz+zx}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(zy+yz+zx\right)=\dfrac{8}{3}\left(xy+yz+zx\right)\)
\(\Rightarrow P\le\dfrac{xy+yz+zx}{2.\dfrac{8}{3}\left(xy+yz+zx\right)}=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)
\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)
Tương tự và nhân vế với vế:
\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)
Mặt khác ta có:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)
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a: Vì \(\dfrac{1}{2}\ne\dfrac{-2}{1}=-2\)
nên hệ phương trình luôn có nghiệm duy nhất
b: \(\left\{{}\begin{matrix}x-2y=4-m\\2x+y=8m+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=4-m\\4x+2y=16m+6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+4x=4-m+16m+6=15m+10\\2x+y=8m+3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=15m+10\\y=8m+3-2x\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3m+2\\y=8m+3-6m-4=2m-1\end{matrix}\right.\)
Đặt \(A=x^2+y^2\)
\(=\left(3m+2\right)^2+\left(2m-1\right)^2\)
\(=9m^2+12m+4+4m^2-4m+1\)
\(=13m^2+8m+5\)
\(=13\left(m^2+\dfrac{8}{13}m+\dfrac{5}{13}\right)\)
\(=13\left(m^2+2\cdot m\cdot\dfrac{4}{13}+\dfrac{16}{169}+\dfrac{49}{169}\right)\)
\(=13\left(m+\dfrac{4}{13}\right)^2+\dfrac{49}{13}>=\dfrac{49}{13}\forall m\)
Dấu '=' xảy ra khi \(m+\dfrac{4}{13}=0\)
=>\(m=-\dfrac{4}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2-2x-4y-4=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2-9=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-2\right)^2=9=0^2+3^2=0^2+\left(-3\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=0\\y-2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=3\\y-2=0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=0\\y-2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-3\\y-2=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow-2\le x\le4\left(y\in R\right)\)
Ta có \(S=3x+4y\)
Mà \(x\ge-2;y\ge-1\Leftrightarrow S\ge3\cdot\left(-2\right)+4\cdot\left(-1\right)=-6-4=-10\)
Vậy GTNN của S là \(-10\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)
Lời giải:
ĐKĐB $\Leftrightarrow (x^2-2x+1)+(y^2-4y+4)-9=0$
$\Leftrightarrow (x-1)^2+(y-2)^2-9=0$
$\Rightarrow (x-1)^2=9-(y-2)^2\leq 9$
$\Rightarrow -3\leq x-1\leq 3$
$\Leftrightarrow -2\leq x\leq 4$
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Đặt $x-1=a; y-2=b$ thì bài toán trở thành:
Cho $a,b$ thực thỏa mãn $a^2+b^2=9$
Tìm min $S=3a+4b+11$
Áp dụng BĐT Bunhiacopxky:
$(3a+4b)^2\leq (a^2+b^2)(3^2+4^2)=9.25$
$\Rightarrow -15\leq 3a+4b\leq 15$
$\Rightarrow 3a+4b\geq -15$
$\Rightarrow S=3a+4b+11\geq -4$
Vậy $S_{\min}=-4$ khi $x=\frac{-4}{5}; y=\frac{-1}{5}$
Lời giải:
Thực hiện khai triển và rút gọn ta có:
$A=3x^2y^2+2x^4+2y^4-(x^2+y^2)=\frac{3}{2}(x^2+y^2)^2+\frac{x^4+y^4}{2}-(x^2+y^2)$
Áp dụng BĐT AM-GM:
\(x^4+y^4\geq 2\sqrt{x^4y^4}=2x^2y^2\)
\(\Rightarrow 2(x^4+y^4)\geq x^4+y^4+2x^2y^2=(x^2+y^2)^2\)
\(\Rightarrow x^4+y^4\geq \frac{(x^2+y^2)^2}{2}\)
\(\Rightarrow A\geq \frac{3}{2}(x^2+y^2)^2+\frac{(x^2+y^2)^2}{4}-(x^2+y^2)\)
Đặt $x^2+y^2=t$
Ta có: $t=x^2+y^2=\frac{1}{2}(x+y)^2+\frac{1}{2}(x-y)^2\geq \frac{(x+y)^2}{2}\geq \frac{1}{2}$ do $x+y\geq 1$
Do đó: \(A\geq \frac{3}{2}t^2+\frac{t^2}{4}-t=\frac{7}{4}t^2-t=(t-\frac{1}{2})(\frac{7}{4}t-\frac{1}{8})-\frac{1}{16}\geq \frac{-1}{16}\) với mọi $t\geq \frac{1}{2}$
Vậy $A_{\min}=\frac{-1}{16}$
Dấu "=" xảy ra khi $x=y=\frac{1}{2}$