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\(\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\frac{\left(a^2+ab+ac+bc\right)\left(b^2+bc+ba+ac\right)}{c^2+ca+cb+ab}}=\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}=a+b\left(a,b,c>0;a+b+c=1\right)\)
Bạn làm tương tự nha
\(\Rightarrow P=a+b+c+a+b+c=2\left(a+b+c\right)=2\)
Trước hết ta chứng minh bài toán quen thuộc:
Cho \(abc=1\) thì \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=1\)
\(VT=\frac{1}{ab+b+1}+\frac{1}{bc+c+abc}+\frac{b}{abc+ab+b}=\frac{1}{ab+b+1}+\frac{1}{c\left(b+1+ab\right)}+\frac{b}{1+ab+b}\)
\(=\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}=\frac{1+ab+b}{ab+b+1}=1\)
\(P=\sum\frac{1}{a^2+2b^2+3}=\sum\frac{1}{a^2+b^2+b^2+1+2}\le\sum\frac{1}{2ab+2b+2}=\frac{1}{2}\sum\frac{1}{ab+b+1}=\frac{1}{2}\)
\(\Rightarrow P_{max}=\frac{1}{2}\) khi \(a=b=c=1\)
\(P=\sum\frac{1}{a^2+1+2\left(b^2+1\right)}\le\sum\frac{1}{2a+4b}=\frac{1}{2}\sum\frac{1}{a+b+b}\le\frac{1}{18}\sum\left(\frac{1}{a}+\frac{2}{b}\right)\)
\(\Rightarrow P\le\frac{1}{18}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{6}.3\sqrt[3]{\frac{1}{abc}}=\frac{1}{2}\)
\(\Rightarrow P_{max}=\frac{1}{2}\) khi \(a=b=c=1\)
\(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)
Tương tự: \(b+ca=\left(a+b\right)\left(b+c\right)\) ; \(c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Ta có: \(P=\Sigma\dfrac{a^2\left(b+1\right)}{a\left(b+1\right)+b}=\Sigma\dfrac{a^2\left(b+1\right)+ab-ab}{a\left(b+1\right)+b}=\Sigma\left(a-\dfrac{ab}{a\left(b+1\right)+b}\right)\)
\(\Rightarrow P=\left(a+b+c\right)-\Sigma\dfrac{ab}{a\left(b+1\right)+b}=3-\Sigma\dfrac{ab}{a\left(b+1\right)+b}\)
Áp dụng BĐT Cauchy \(\Rightarrow a\left(b+1\right)+b=ab+b+a\ge3\sqrt[3]{a^2b^2}\)
\(\Rightarrow P\ge3-\Sigma\dfrac{ab}{\sqrt[3]{a^2b^2}}=3-\Sigma\dfrac{\sqrt[3]{ab}}{3}\)
mà \(\sqrt[3]{ab}=\sqrt[3]{a.b.1}\le\dfrac{a+b+1}{3}\)
\(3-\Sigma\dfrac{\sqrt[3]{ab}}{3}=3-\dfrac{\sqrt[3]{ab}+\sqrt[3]{bc}+\sqrt[3]{ac}}{3}\ge3-\dfrac{\dfrac{2\left(a+b+c\right)+3}{3}}{3}=3-1=2\)
\(\Rightarrow P\ge2\) \(\Rightarrow MinP=2\) khi a = b = c =1
Lời giải khác:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{a^2(b+1)}{a+b+ab}+\frac{b^2(c+1)}{b+c+bc}+\frac{c^2(a+1)}{c+a+ac}\)\(=\frac{a^2}{\frac{a+b+ab}{b+1}}+\frac{b^2}{\frac{b+c+bc}{c+1}}+\frac{c^2}{\frac{c+a+ca}{a+1}}\)
\(\geq \frac{(a+b+c)^2}{\frac{(a+1)(b+1)-1}{b+1}+\frac{(b+1)(c+1)-1}{c+1}+\frac{(c+1)(a+1)-1}{a+1}}\)
\(\Leftrightarrow P\geq \frac{9}{a+b+c+3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}=\frac{9}{6-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq \frac{9}{a+1+b+1+c+1}=\frac{9}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)
Do đó: \(6-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\leq 6-\frac{3}{2}=\frac{9}{2}\)
\(\Rightarrow P\geq \frac{9}{\frac{9}{2}}=2\)
Vậy P min là 2
Dấu bằng xảy ra khi \(a=b=c=1\)
Lời giải:
Do $ab+bc+ac=5$ nên:
\(a^2+5=a^2+ab+bc+ac=(a+b)(a+c)\)
\(b^2+5=b^2+ab+bc+ac=(b+c)(b+a)\)
\(c^2+5=c^2+ab+bc+ac=(c+a)(c+b)\)
Do đó:
\(A=a\sqrt{\frac{(b+c)(b+a)(c+a)(c+b)}{(a+b)(a+c)}}+b\sqrt{\frac{(a+b)(a+c)(c+a)(c+b)}{(b+c)(b+a)}}+c\sqrt{\frac{(a+b)(a+c)(b+c)(b+a)}{(c+a)(c+b)}}\)
\(=a\sqrt{(b+c)^2}+b\sqrt{(c+a)^2}+c\sqrt{(a+b)^2}=a(b+c)+b(c+a)+c(a+b)\)
\(=2(ab+bc+ac)=2.5=10\)
Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)
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Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)\Rightarrow x+y+z=3\)
\(K=\dfrac{z^3}{x^2+z^2}+\dfrac{x^3}{x^2+y^2}+\dfrac{y^3}{y^2+z^2}\)
Ta chứng minh BĐT phụ sau: \(\dfrac{x^3}{x^2+y^2}\ge\dfrac{2x-y}{2}\)
Thật vậy, BĐT tương đương:
\(2x^3\ge2x^3-x^2y+2xy^2-y^3\)
\(\Leftrightarrow y\left(x-y\right)^2\ge0\) (đúng)
Tương tự: \(\dfrac{y^3}{y^2+z^2}\ge\dfrac{2y-z}{2}\) ; \(\dfrac{z^3}{z^2+x^2}\ge\dfrac{2z-x}{2}\)
Cộng vế với vế:
\(K\ge\dfrac{x+y+z}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=\dfrac{1}{3}\)
Không biết cách làm đúng k nữa :D
Đặt: \(\hept{\begin{cases}a+bc=7^x\\b+ac=7^y\end{cases}}\)
TH1: Nếu \(7^x=7^y\)khi đó: n chẵn
\(\Leftrightarrow a+bc=b+ac\)
\(\Leftrightarrow\left(a-b\right)\left(1-c\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\c=1\end{cases}}\)
TH2:Nếu: \(7^x>7^y\)(*)
\(\Leftrightarrow a+bc>b+ac\)
\(\Leftrightarrow\left(a-b\right)\left(1-c\right)>0\)
\(\hept{\begin{cases}a>b\\c< 1\end{cases}\left(ktm\right)}\)hoặc: \(\hept{\begin{cases}a< b\\c>1\end{cases}\left(tm\right)}\)(1)
Đồng thời phải thỏa mãn điều kiện: \(a+bc⋮b+ac\)
\(\Leftrightarrow\left(a-b\right)\left(1-c\right)⋮b+ac\)
\(\Leftrightarrow\orbr{\begin{cases}a-b⋮b+ac\\1-c⋮b+ac\end{cases}\Leftrightarrow}\orbr{\begin{cases}a+ac⋮b+ac\\a\left(1-c\right)⋮b+ac\end{cases}\Leftrightarrow\orbr{\begin{cases}a+ac⋮b+ac\\a+b⋮b+ac\end{cases}}}\)(2)
Vì a,b,c thuộc N* nên:
\(\left(1\right)\Leftrightarrow\hept{\begin{cases}a+ac< b+ac\\ac+b>a+b\end{cases}}\)
Mặt khác: \(a+ac;a+b\ne0\)
Nên (2) sai
Dẫn đến (*) sai
Tương tự xét: \(7^x< 7^y\)(loại)
Vậy n chẵn
k cho tui