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Lời giải:
$P=a^3b^3+1+1+\frac{1}{a^3b^3}$
$=(ab)^3+\frac{1}{(ab)^3}+2$
Áp dụng BĐT Cô-si:
$(ab)^3+\frac{1}{4096(ab)^3}\geq 2\sqrt{(ab)^3.\frac{1}{4096(ab)^3}}=\frac{1}{32}(1)$
$ab\leq \frac{(a+b)^2}{4}=\frac{1}{4}$
$\Rightarrow (ab)^3\leq \frac{1}{64}$
$\Rightarrow \frac{4095}{4096(ab)^3}\geq \frac{4095}{64}(2)$
Từ $(1);(2)$ suy ra:
$P\geq \frac{1}{32}+\frac{4095}{64}+2=\frac{4225}{64}$
Vậy $P_{\min}=\frac{4225}{64}$
Giá trị này đạt tại $a=b=\frac{1}{2}$
Ta có: \(2\left(b^2+bc+c^2\right)=2b^2+2c^2+2bc\le2b^2+2c^2+b^2+c^2=3\left(b^2+c^2\right)\Rightarrow b^2+c^2\le3-a^2\Rightarrow a^2+b^2+c^2\le3\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\).
Áp dụng bđt Schwars ta có:
\(T\ge a+b+c+\dfrac{18}{a+b+c}=\left(a+b+c+\dfrac{9}{a+b+c}\right)+\dfrac{9}{a+b+c}\ge2\sqrt{9}+\dfrac{9}{3}=9\).
Đẳng thức xảy ra khi a = b = c = 1.
Lời giải:
\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\)
Xét:
\(x^4+8xy^3-(x^2+2y^2)^2=8xy^3-4y^4-4x^2y^2\)
\(=-4y^2(x^2-2xy+y^2)=-4y^2(x-y)^2\leq 0\)
\(\Rightarrow x^4+8xy^3\leq (x^2+2y^2)^2\)
\(\Rightarrow \frac{x^2}{\sqrt{x^4+8xy^3}}\geq \frac{x^2}{x^2+2y^2}(*)\)
Mặt khác:
\(y^4+y(x+y)^3-(x^2+2y^2)^2=x^3y+3xy^3-2y^4-x^4-x^2y^2\)
\(=x^3(y-x)+3y^3(x-y)+y^4-x^2y^2\)
\(=x^3(y-x)+3y^3(x-y)+y^2(y-x)(y+x)\)
\(=(y-x)(x^3-2y^3+xy^2)\)
\(=(y-x)[(x-y)(x^2+xy+y^2)+y^2(x-y)]\)
\(=-(x-y)^2(x^2+xy+2y^2)\leq 0\)
\(\Rightarrow y^4+y(x+y)^3\leq (x^2+2y^2)^2\Rightarrow \frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\geq \frac{2y^2}{x^2+2y^2}(**)\)
Từ $(*); (**)\Rightarrow A\geq 1$
\(GT\Leftrightarrow a^2+b^2-2ab=a+b+2\)
\(\Leftrightarrow a^2+a+b^2+b=2\left(ab+a+b+1\right)\)
\(\Leftrightarrow a\left(a+1\right)+b\left(b+1\right)=2\left(a+1\right)\left(b+1\right)\)
\(\Leftrightarrow\dfrac{a}{b+1}+\dfrac{b}{a+1}=2\)
Đặt \(\left(\dfrac{a}{b+1};\dfrac{b}{a+1}\right)=\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}x;y\ge0\\x+y=2\end{matrix}\right.\)
\(\Rightarrow0\le xy\le1\)
\(P=\left(1+x^3\right)\left(1+y^3\right)=1+x^3+y^3+x^3y^3\)
\(P=1+\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3\)
\(P=\left(xy\right)^3-6xy+9=xy\left[\left(xy\right)^2-6\right]+9\le9\)
Dấu "=" xảy ra khi \(xy=0\Leftrightarrow\left(a;b\right)=\left(0;2\right);\left(2;0\right)\)
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^
ĐKXĐ: \(abc\ne0\)
\(a^3+b^3+3ab\left(a+b\right)+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
TH1: \(a+b+c=0\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
TH2: \(a=b=c\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
\(A=\dfrac{\left(a+b+c+a\right)\left(a+b+c+b\right)\left(a+b+c+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(A\ge\dfrac{2\sqrt{\left(a+b\right)\left(a+c\right)}.2\sqrt{\left(a+b\right)\left(b+c\right)}.2\sqrt{\left(a+c\right)\left(b+c\right)}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
\(A=a^3b^3+\dfrac{1}{a^3b^3}+2=a^3b^3+\dfrac{1}{2^{12}.a^3b^3}+\dfrac{2^{12}-1}{2^{12}a^3b^3}+2\)
\(A\ge2\sqrt{\dfrac{a^3b^3}{2^{12}.a^3b^3}}+\dfrac{2^{12}-1}{2^{12}.\left(\dfrac{a+b}{2}\right)^6}+2=\dfrac{2}{2^6}+\dfrac{2^{12}-1}{2^6}+2=\dfrac{2^{12}+1}{2^6}+2\) (casio)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)