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Đặt \(x=2a\)và \(y=2b\)suy ra \(\hept{\begin{cases}x>0\\y>0\\x+y\le2\end{cases}}\)
Suy ra : \(A=\frac{x}{y+2}+\frac{y}{x+2}+\frac{2}{x+y}\)
\(\Rightarrow A=\frac{x^2}{xy+2x}+\frac{y^2}{xy+2y}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(xy+x+y\right)}+\frac{2}{x+y}\)
\(\Rightarrow A\ge\frac{\left(x+y\right)^2}{2\left(\frac{\left(x+y\right)^2}{4}+\left(x+y\right)\right)}+\frac{2}{x+y}\)
Đặt \(t=x+y\)( \(0< t\le2\))
Suy ra :
\(\Rightarrow A\ge\frac{t^2}{\frac{t^2}{2}+2t}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{2}{t}\)
\(\Rightarrow A\ge\frac{2t}{t+4}+\frac{4}{3}.\frac{1}{t}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{2t}{t+4}.\frac{4}{3}.\frac{1}{t}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3\left(t+4\right)}}+\frac{2}{3}.\frac{1}{t}\)
\(\Rightarrow A\ge2\sqrt{\frac{8}{3.\left(2+4\right)}}+\frac{2}{3}.\frac{1}{2}=\frac{5}{3}\)
"=" xảy ra khi \(x=y=\frac{1}{2}\)
1) Áp dụng BĐT Cauchy-Schwarz, ta có:
\(VT=\dfrac{9}{3\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{16}{\left(a+b+c\right)^2+ab+bc+ca}=\dfrac{16}{1+ab+bc+ca}\ge\dfrac{16}{1+\dfrac{\left(a+b+c\right)^2}{3}}=\dfrac{16}{1+\dfrac{1}{3}}=12\)
Lưu ý: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Đẳng thức xảy ra khi a=b=c=1/3
Post lại :v
1) Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT=\dfrac{1}{ab+bc+ca}+\dfrac{4}{2\left(ab+bc+ca\right)}+\dfrac{1}{a^2+b^2+c^2}\)
\(VT\ge\dfrac{3}{\left(a+b+c\right)^2}+\dfrac{\left(2+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)
\(VT\ge3+\dfrac{9}{\left(a+b+c\right)^2}=3+9=12\)(đpcm)
Đảng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)
2) Áp dụng BĐT Cauchy-Schwarz, ta có:
\(VT=\dfrac{\dfrac{2}{3}}{ab}+\dfrac{\dfrac{1}{3}}{ab}+\dfrac{3}{a^2+b^2+ab}\)
\(VT\ge\dfrac{\dfrac{2}{3}}{\dfrac{\left(a+b\right)^2}{4}}+\dfrac{\left(\dfrac{1}{\sqrt{3}}+\sqrt{3}\right)^2}{a^2+b^2+ab+ab}\)
\(VT\ge\dfrac{\dfrac{2}{3}}{\dfrac{1}{4}}+\dfrac{\dfrac{16}{3}}{\left(a+b\right)^2}=\dfrac{8}{3}+\dfrac{16}{3}=\dfrac{24}{3}=8\)(đpcm)
Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)
Đang rảnh, làm luôn\(A=\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}=\dfrac{1}{2}\left[\left(\dfrac{a}{bc}+\dfrac{b}{ca}\right)+\left(\dfrac{b}{ca}+\dfrac{c}{ab}\right)+\left(\dfrac{c}{ab}+\dfrac{a}{bc}\right)\right]\ge\dfrac{1}{2}\left(\dfrac{2}{c}+\dfrac{2}{a}+\dfrac{2}{b}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 2
Lời giải:
\(a+b=ab\Rightarrow \frac{1}{a}+\frac{1}{b}=1\)
Đặt \(\left(\frac{1}{a}, \frac{1}{b}\right)=(x,y)\) thì bài toán trở thành:
Cho $x,y>0$ thỏa mãn $x+y=1$. Tìm GTNN của biểu thức:
\(P=\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\)
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Áp dụng BĐT Cauchy-Schwarz, AM-GM:
\(\frac{x^2}{2x+1}+\frac{y^2}{2y+1}\geq \frac{(x+y)^2}{2x+1+2y+1}=\frac{1}{2+2}=\frac{1}{4}\)
\((x^2+1)(y^2+1)\geq (xy+1)^2\Rightarrow \frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\geq \frac{xy+1}{xy}=1+\frac{1}{xy}\)
\(\geq 1+\frac{1}{\frac{(x+y)^2}{4}}=5\)
\(\Rightarrow P=\frac{x^2}{2x+1}+\frac{y^2}{2y+1}+\frac{\sqrt{(x^2+1)(y^2+1)}}{xy}\geq \frac{1}{4}+5=\frac{21}{4}\)
Vậy \(P_{\min}=\frac{21}{4}\Leftrightarrow x=y=\frac{1}{2}\Leftrightarrow a=b=2\)
giải tạm 1 bài z -,-
2) Cauchy-Schwarz dạng Engel :
\(A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}=\dfrac{6}{2}=3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)
Chúc bạn học tốt ~
4/ Ta có: \(6=a+b+c+ab+bc+ca\ge3\left(\sqrt[3]{\left(abc\right)^2}+\sqrt[3]{abc}\right)\)
Đặt \(\sqrt[3]{abc}=t\Rightarrow t^2+t\le2\Rightarrow t\le1\Rightarrow t^3=C=abc\le1\)
Vậy...
5/ \(D\le\left(\frac{a+b+c}{3}\right)^3.\left[\frac{2\left(a+b+c\right)}{3}\right]^3=\frac{512}{729}\)
Vậy ...
P/s: Em không chắc
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)^2\)
\(\Rightarrow \frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+bc+ac}{abc}=ab+bc+ac\)
Do đó:
\(P\geq ab+bc+ac+\frac{9}{2(a+b+c)}\)
Áp dụng BĐT AM-GM:
\(ab+bc+ac+\frac{9}{2(a+b+c)}=\frac{ab+bc+ac}{2}+\frac{ab+bc+ac}{2}+\frac{9}{2(a+b+c)}\geq 3\sqrt[3]{\frac{9(ab+bc+ac)^2}{8(a+b+c)}}\)
Theo một kết quả quen thuộc của BĐT AM-GM:
\((ab+bc+ac)^2\geq 3abc(a+b+c)\)
Thay \(abc=1\Rightarrow (ab+bc+ac)^2\geq 3(a+b+c)\)
Do đó: \(P\geq ab+bc+ac+\frac{9}{2(a+b+c)}\geq 3\sqrt[3]{\frac{27}{8}}=\frac{9}{2}\)
Vậy \(P_{\min}=\frac{9}{2}\Leftrightarrow a=b=c=1\)
ap dung bdt cosi ta co : \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge3\sqrt[3]{\dfrac{abc}{\left(abc\right)^2}}=3\) (1)
ta lai co \(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9\left(a+b+c\right)}{2\left(a+b+c\right)^2}\ge\dfrac{9.3}{2.3^2}=\dfrac{3}{2}\) (2)
tu (1) vs (2) \(\Rightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{9}{2\left(a+b+c\right)}\ge3+\dfrac{3}{2}=\dfrac{9}{2}\)
dau "=" xay ra khi \(a=b=c=1\)
xl ! may mk bi hu nen khong viet dau dc bn thong cam
Câu 4:
Để C chia hết cho D thì \(x^4+a⋮x^2+4\)
\(\Leftrightarrow x^4-16+a+16⋮x^2+4\)
=>a+16=0
hay a=-16
\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)
Áp dụng bđt Cauchy-Schwarz dạng Engel có:
\(A\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\ge\dfrac{4}{\left(a+b\right)^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{2}}\ge\dfrac{4}{1}+\dfrac{1}{\dfrac{1}{2}}=6\)
Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)
Vậy GTNN của A=6
Thiếu đề ko e?