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Với 2 số thực x,y>0, ta có:
\(x^3+y^3-x^2y-xy^2=\left(x+y\right)\left(x-y\right)^2\ge0\). Dấu bằng xảy ra \(\Leftrightarrow x=y\).
Do đó: \(x^3+y^3\ge x^2y+xy^2\Leftrightarrow4x^3+4y^3\ge\left(x+y\right)^3\Leftrightarrow x+y\le\sqrt[3]{4x^3+4y^3}\)Áp dụng bđt vừa cm, ta có: \(S=\sqrt[3]{2a+b}+\sqrt[3]{2b+c}+\sqrt[3]{2c+d}+\sqrt[3]{2d+a}\le\sqrt[3]{8a+12b+4c}+\sqrt[3]{8c+12d+4a}\le\sqrt[3]{48a+48b+48c+48d}=\sqrt[3]{48}\)(vì a+b+c+d=1)
Dấu bằng xảy ra\(\Leftrightarrow a=b=c=d=\dfrac{1}{4}\)(vì a+b+c+d=1)
Bn ơi 3x3 + 3y3 vào cả 2 vế thì 4x3 + 4y3 > 3x3 + 3y3 + x2y + xy2 k phải là (x + y)3
1) hệ <=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}\left(\sqrt[3]{x-1}+\sqrt[3]{y+1}\right)=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+y+3\sqrt[3]{xy}=1\\x+y+3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=1\end{matrix}\right.\)
trừ vế theo vế => \(3\sqrt[3]{xy}-3\sqrt[3]{\left(x-1\right)\left(y+1\right)}=0\)
<=> xy=(x-1)(y-1) <=> x-y=1=> \(\left\{{}\begin{matrix}\sqrt[3]{x}+\sqrt[3]{y}=1\\x-y=1\end{matrix}\right.\)
đặt \(\sqrt[3]{x}=a;\sqrt[3]{y}=b\)
hpt <=> \(\left\{{}\begin{matrix}a+b=1\\a^3-b^3=1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}b=1-a\\2a^3-3a^2+3a-2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}b=1-a\\\left(a-1\right)\left(2a^2-a+2\right)=0\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
p/s: cách làm khá dài ,có ai có cách khác thì làm luôn cho mik exp :v )
Bài 2:Thêm đk a, b, c không âm.
Theo Bunhiacopxki: \(Q^2\le3\left[2\left(a+b+c\right)+ab+bc+ca\right]\)
\(\le3\left(2.2+\frac{\left(a+b+c\right)^2}{3}\right)=16\Rightarrow Q\le4\)
Đẳng thức xảy ra khi a = b = c = 2/3
Vậy..
tth Tui có cách khác ông nè:)
\(Q=\sqrt{2a+bc}+\sqrt{2b+ac}+\sqrt{2c+ab}\)
\(Q=\sqrt{\left(a+b+c\right)\cdot a+bc}+\sqrt{\left(a+b+c\right)\cdot b+ac}+\sqrt{\left(a+b+c\right)\cdot c+ab}\)
\(Q=\sqrt{a^2+ab+ac+bc}+\sqrt{ab+b^2+bc+ac}+\sqrt{ac+bc+c^2+ab}\)
\(Q=\sqrt{\left(a+c\right)\left(a+b\right)}+\sqrt{\left(b+c\right)\left(b+a\right)}+\sqrt{\left(c+a\right)\left(a+b\right)}\)
Áp dụng bất đẳng thức AM-GM ngược dấu ta có:
\(Q\le\frac{a+c+a+b}{2}+\frac{b+c+b+a}{2}+\frac{c+a+a+b}{2}\)
\(=\frac{4\left(a+b+c\right)}{2}=4\)
Vậy \(Q_{max}=4\Leftrightarrow a=b=c=\frac{2}{3}\)
Đề sai phải là \(\sqrt{2b^2+bc+2c^2}\)
\(\sqrt{2a^2+ab+2b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\ge\sqrt{\frac{5}{4}}\left(a+b\right)\)
CMTT, có: \(\sqrt{2b^2+bc+2c^2}\ge\sqrt{\frac{5}{4}}\left(b+c\right)\)
\(\sqrt{2c^2+ca+2a^2}\ge\sqrt{\frac{5}{4}}\left(c+a\right)\)
\(\Rightarrow P\ge\sqrt{5}\left(a+b+c\right)\ge\frac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\frac{\sqrt{5}}{3}\)
Dấu "=" xảy ra khi a=b=c=\(\frac{1}{9}\)
Bài 1:
\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có:
\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)
Cộng theo vế 3 BĐT trên ta có:
\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)
Đẳng thức xảy ra khi \(a=b=c\)
Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2
Bài 2/
\(\frac{bc}{a^2b+a^2c}+\frac{ca}{b^2c+b^2a}+\frac{ab}{c^2a+c^2b}\)
\(=\frac{b^2c^2}{a^2b^2c+a^2c^2b}+\frac{c^2a^2}{b^2c^2a+b^2a^2c}+\frac{a^2b^2}{c^2a^2b+c^2b^2a}\)
\(=\frac{b^2c^2}{ab+ac}+\frac{c^2a^2}{bc+ba}+\frac{a^2b^2}{ca+cb}\)
\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)
\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)
Dấu = xảy ra khi \(a=b=c=1\)
\(\frac{1}{2}\left(a+b\right)^2\le a^2+b^2=ab\left(a+b\right)+ab\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{2}\left(a+b\right)^2\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{4}\left(a+b\right)^2\le ab\left(a+b\right)\Rightarrow a+b\le4ab\)
\(\Rightarrow\frac{a+b}{ab}\le4\)
\(P=\frac{\sqrt{b\left(a+b\right)}}{ab}+\frac{\sqrt{a\left(a+b\right)}}{ab}=\frac{1}{2\sqrt{2}}\left(\frac{2\sqrt{2b\left(a+b\right)}+2\sqrt{2a\left(a+b\right)}}{ab}\right)\)
\(P\le\frac{1}{2\sqrt{2}}\left(\frac{2b+a+b+2a+a+b}{ab}\right)=\sqrt{2}\left(\frac{a+b}{ab}\right)\le4\sqrt{2}\)
\(P_{max}=4\sqrt{2}\) khi \(a=b=\frac{1}{2}\)
Ta có : \(\sqrt{2a^2+ab+b^2}=\sqrt{\frac{5}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}\)
Vì \(\frac{3}{4}\left(a-b\right)^2\ge0\forall a;b\Rightarrow\sqrt{2a^2+ab+b^2}\ge\sqrt{\frac{5}{4}}\left(a+b\right)\)( 1 )
Tương tự , ta có : \(\sqrt{2b^2+bc+c^2}\ge\sqrt{\frac{5}{4}}\left(b+c\right);\sqrt{2c^2+ac+a^2}\ge\sqrt{\frac{5}{4}}\left(a+c\right)\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow P\ge\sqrt{\frac{5}{4}}.2\left(a+b+c\right)=\sqrt{5}\left(a+b+c\right)\)
Áp dụng BĐT phụ \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) , ta có :
\(P\ge\sqrt{5}.\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{3}=\frac{\sqrt{5}}{3}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=\frac{1}{9}\)
Áp dụng BĐT Bu-nhi-a-cốp-xki ta có:
\(S^2=\left(1.\sqrt{1+2a}+1.\sqrt{1+2b}\right)^2\le\left(1^2+1^2\right)\left(1+2a+1+2b\right)\)\(=4+4\left(a+b\right)\)
Áp dụng tiếp BĐT Bu-nhi-a-cốp-xki ta có \(\left(a+b\right)^2\le2\left(a^2+b^2\right)=2\)\(\Rightarrow a+b\le\sqrt{2}\)\(\Rightarrow S^2\le4+4\sqrt{2}\Rightarrow S\le2\sqrt{1+\sqrt{2}}\).Xảy ra đẳng thức khi và chỉ khi \(a=b=\frac{\sqrt{2}}{2}\)