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Tham khảo lời giải:
Đặt (xyz;yzx;xzy)=(a,b,c)(xyz;yzx;xzy)=(a,b,c)
⇒⎧⎪⎨⎪⎩y2=abx2=acz2=bc⇒{y2=abx2=acz2=bc
Bài toán trở thành: Cho $a,b,c>0$ thỏa mãn ab+bc+ac=1ab+bc+ac=1
Tìm min $S=a+b+c$
Theo hệ quả quen thuộc của BĐT Cauchy: (a+b+c)2≥3(ab+bc+ac)(a+b+c)2≥3(ab+bc+ac)
⇒S=√(a+b+c)2≥√3(ab+bc+ac)=√3⇒S=(a+b+c)2≥3(ab+bc+ac)=3
Vậy Smin=√3⇔a=b=c=13⇔x=y=z=1√3
\(x-3=y\left(x+1\right)\Rightarrow y=\frac{x-3}{x+1}\)
\(A=x^2+\left(\frac{x-3}{x+1}\right)^2=x^2+\left(1-\frac{4}{x+1}\right)^2=x^2+1-\frac{8}{x+1}+\frac{16}{\left(x+1\right)^2}\)
\(=\left(x+1\right)^2-2x-\frac{8}{x+1}+\frac{16}{\left(x+1\right)^2}=\left(x+1\right)^2+\frac{16}{\left(x+1\right)^2}-2\left(x+1+\frac{4}{x+1}\right)+2\)
Đặt \(x+1+\frac{4}{x+1}=a\Rightarrow a^2=\left(x+1\right)^2+\frac{16}{\left(x+1\right)^2}+8\) (\(\left|a\right|\ge4\))
\(\Rightarrow A=a^2-8-2a+2=a^2-2a-6\)
- Nếu \(a\le-4\Rightarrow A=\left(a+4\right)^2-10a-22\ge-10a-22\ge40-22=18\)
- Nếu \(a\ge4\Rightarrow A=\left(a-4\right)^2+6a-22\ge6a-22\ge24-22=2\)
\(\Rightarrow A_{min}=2\) khi \(a=4\Rightarrow x+1+\frac{4}{x+1}=4\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)
Dấu = xảy ra khi x=y=2
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
Chia cả tử và mẫu của phân số thứ 3 cho xy
Trần Anh Thơ
\(B=\frac{x}{y}+\frac{y}{x}-1+\frac{1}{\frac{x}{y}+\frac{y}{x}-1}+1\ge2\sqrt{\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{1}{\frac{x}{y}+\frac{y}{x}-1}\right)}+1=3\)
\(B_{min}=3\) khi \(\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x=y\)
\(P=\frac{5}{x^2+y^2}+\frac{5}{2xy}+\frac{1}{2xy}=5\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\)
\(P\ge\frac{5.4}{x^2+y^2+2xy}+\frac{2}{\left(x+y\right)^2}=\frac{22}{\left(x+y\right)^2}=\frac{22}{9}\)
\(\Rightarrow P_{min}=\frac{22}{9}\) khi \(x=y=\frac{3}{2}\)
\(M=x^2+\frac{y^2}{4}+\frac{1}{4}-xy-x+\frac{y}{2}+\frac{3y^2}{4}+\frac{y}{2}+\frac{3}{4}\)
\(M=\left(x-\frac{y}{2}-\frac{1}{2}\right)^2+\frac{3}{4}\left(y+1\right)^2\ge0\)
\(\Rightarrow M_{min}=0\) khi \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)