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Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
https://hoc24.vn/cau-hoi/cho-abc-0-thoa-man-abbcca3-tim-gia-tri-nho-nhat-cua-pdfrac13a1b2dfrac13b1c2dfrac13c1a2.6181078378966
Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
Ta có BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.3=9\)
\(\Rightarrow a+b+c\ge3\)
Phân tích và áp dụng BĐT AM-GM:
\(\dfrac{1+3a}{1+b^2}=\dfrac{1}{1+b^2}+\dfrac{3a}{1+b^2}=\left(1-\dfrac{b^2}{1+b^2}\right)+\left(3a-\dfrac{3ab^2}{1+b^2}\right)\ge\left(1-\dfrac{b^2}{2b}\right)+\left(3a-\dfrac{3ab^2}{2b}\right)=\left(1-\dfrac{b}{2}\right)+\left(3a-\dfrac{3}{2}ab\right)\)
Tương tự:
\(\dfrac{1+3b}{1+c^2}\ge\left(1-\dfrac{c}{2}\right)+\left(3b-\dfrac{3}{2}bc\right)\)
\(\dfrac{1+3c}{1+a^2}\ge\left(1-\dfrac{a}{2}\right)+\left(3c-\dfrac{3}{2}ca\right)\)
Cộng các vế của các BĐT ta được:
\(P\ge3-\dfrac{1}{2}\left(a+b+c\right)+3\left(a+b+c\right)-\dfrac{3}{2}\left(ab+bc+ca\right)=3+\dfrac{5}{2}\left(a+b+c\right)-\dfrac{3}{2}.3\ge3+\dfrac{5}{2}.3-\dfrac{9}{2}=6\)
\(P=6\Leftrightarrow a=b=c=1\)
Vậy \(P_{min}=6\)
Áp dụng BĐT: \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\) ( Câu hỏi của ZoZ - Kudo vs Conan - ZoZ - Toán lớp 9 | Học trực tuyến)
\(\Rightarrow\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Áp dụng vào, ta có:
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{ab}{9}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)\)\(\dfrac{bc}{b+3c+2a}=\dfrac{bc}{\left(a+c\right)+\left(a+b\right)+2c}\le\dfrac{9}{bc}\left(\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{2c}\right)\)\(\dfrac{ca}{c+3a+2b}=\dfrac{ca}{\left(c+b\right)+\left(b+a\right)+2a}\le\dfrac{ca}{9}\left(\dfrac{1}{c+b}+\dfrac{1}{b+a}+\dfrac{1}{2a}\right)\)
Cộng vế theo vế BĐT, ta được:
\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}\right)+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{9}\left[\dfrac{c\left(a+b\right)}{a+b}+\dfrac{b\left(c+a\right)}{a+c}+\dfrac{a\left(b+c\right)}{b+c}\right]+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{9}\left(a+b+c\right)+\dfrac{1}{18}\left(a+b+c\right)\)
\(P\le\dfrac{1}{6}\left(a+b+c\right)\) \(=\dfrac{1}{6}.6=1\)
\(\Rightarrow Max_P=1\Leftrightarrow a=b=c=2\)
Áp dụng bất đẳng thức Svác xơ ngược ta có
\(\frac{1}{2a+3b+3c}=\frac{1}{a+b+a+c+2\left(b+c\right)}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{2}{b+c}\right)\)
tương tự mấy cái kia rồi cộng vào
Giải câu 1 thôi câu 2 không hứng lắm:
\(P=\dfrac{1}{2a+3b+c+6}+\dfrac{1}{2b+3c+a+6}+\dfrac{1}{2c+3a+b+6}\)
Ta có:
\(\dfrac{1}{2a+3b+c+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{b+2}\right)=\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{2}{b+2}\right)\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{1}{2b+3c+a+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{b+2}+\dfrac{2}{c+2}\right)\left(2\right)\\\dfrac{1}{2c+3a+b+6}\le\dfrac{1}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{c+2}+\dfrac{2}{a+2}\right)\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được:
\(P\le\dfrac{3}{16}\left(\dfrac{1}{a+b+c}+\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)
\(\le\dfrac{3}{16.3\sqrt[3]{abc}}+\dfrac{3}{16}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)
\(=\dfrac{1}{16}+\dfrac{3}{16}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\left(4\right)\)
Giờ ta tính Max của \(Q=\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)\)
Vì \(abc=1\) nên không mất tính tổng quát ta giả sử \(\left\{{}\begin{matrix}ab\le1\\c\ge1\end{matrix}\right.\)
Ta có: \(Q=\dfrac{1}{2}.\left(\dfrac{1}{\dfrac{a}{2}+2}+\dfrac{1}{\dfrac{b}{2}+2}\right)+\dfrac{1}{c+2}\)
Ta có bổ đề: Với \(x,y>0;xy\le1\) thì
\(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\le\dfrac{2}{xy+1}\)
Áp dụng vào bài toán ta được:
\(Q\le\dfrac{2}{1+\dfrac{\sqrt{ab}}{2}}+\dfrac{1}{c+2}=\dfrac{2\sqrt{c}}{2\sqrt{c}+1}+\dfrac{1}{c+2}\)
Xét hàm số \(f\left(\sqrt{c}\right)=\dfrac{2\sqrt{c}}{2\sqrt{c}+1}+\dfrac{1}{c+2}\) với \(\sqrt{c}\ge1\) thì hàm số \(f\left(\sqrt{c}\right)\) nghịch biến. Vậy Q đạt GTLN khi c bé nhất.
\(\Rightarrow Q\le f\left(1\right)=1\left(2\right)\)
Từ (4) và (5) ta suy ra
\(P\le\dfrac{1}{16}+\dfrac{3}{16}.1=\dfrac{1}{4}\)
Vậy GTLN là \(P=\dfrac{1}{4}\) đạt được khi \(a=b=c=1\)
2) A = n3 - n2 + n - 1
A = n2(n - 1) + (n - 1)
A = (n - 1)(n2 + 1)
Để A nguyên tố thì n > 1
=> n2 + 1 > 1
Mà A = (n - 1)(n2 + 1) là số nguyên tố, chỉ gồm 2 ước là 1 và chính nó
Nên A = n2 + 1; n - 1 = 1
=> n = 2 (TM)
b) n5 - n + 2
= n(n4 - 1) + 2
= n(n2 - 1)(n2 + 1) + 2
= n(n - 1)(n + 1)(n2 + 1) + 2
n(n - 1)(n + 1) là tích 3 số nguyên liên tiếp do n \(\in N\) nên n(n - 1)(n + 1) chia hết cho 3
=> n(n - 1)(n + 1)(n2 + 1) + 2 chia 3 dư 2, không là số chính phương
Vậy ...
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
Theo BĐT Bu nhi a cốp xki ta có :
\(\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\ge16\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{16}{a+b+c+d}\)
Áp dụng vào bài toán ta có :
\(\dfrac{1}{3a+3b+2c}=\dfrac{1}{16}.\dfrac{16}{\left(a+b\right)+\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3b+3c+2a}=\dfrac{1}{16}.\dfrac{16}{\left(b+c\right)+\left(b+c\right)+\left(a+b\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{b+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3c+3a+2b}=\dfrac{1}{16}.\dfrac{16}{\left(c+a\right)+\left(c+a\right)+\left(a+b\right)+\left(b+c\right)}\le\dfrac{1}{16}\left(\dfrac{1}{c+a}+\dfrac{1}{c+a}+\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{3a+3b+2c}+\dfrac{1}{3b+3c+2a}+\dfrac{1}{3c+3a+2b}\le\dfrac{1}{16}\left(\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\right)=\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}.6=\dfrac{3}{2}\)
Vậy GTLN của A là \(\dfrac{3}{2}\) . Dấu \("="\) xảy ra khi \(a=b=c=\dfrac{1}{4}\)
GIÚP MIK NHANH NHANH NHA MẤY CẬU!