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ta có :\(a^2-ab+b^2=\left(a+b\right)^2-3ab\ge\left(a+b\right)^2-\dfrac{3}{4}\left(a+b\right)^2=\dfrac{1}{4}\left(a+b\right)^2\)(theo BĐT AM-GM)
\(\Rightarrow P\ge\sum\dfrac{a+b}{2\sqrt{ab+1}}\)
ÁP dụng BĐT AM-GM:
\(\dfrac{a+b}{2\sqrt{ab+1}}+\dfrac{b+c}{2\sqrt{bc+1}}+\dfrac{c+a}{2\sqrt{ca+1}}\ge3\sqrt[3]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}=\dfrac{3}{2}.\dfrac{1}{\sqrt[3]{\sqrt{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}}}\)
Mà \(\sqrt[3]{\left(ab+1\right)\left(bc+1\right)\left(ca+1\right)}\le\dfrac{1}{3}\left(ab+bc+ca+3\right)\)
\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\left(ab+bc+ca+3\right)}}\)(*)
ta liên tưởng đến BĐT phụ:\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
Cm: phân tích :\(VT=xy\left(x+y\right)+yz\left(y+z\right)+zx\left(x+z\right)+2xyz\)
\(=xy\left(x+y\right)+yz\left(y+z\right)+xz\left(z+x\right)+3xyz-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
mà \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)
nên \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
Áp dụng:
\(1=\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\dfrac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
mặt khác,theo AM-GM,dễ dàng chứng minh được \(a+b+c\ge\dfrac{3}{2}\)
nên \(1\ge\dfrac{8}{9}.\dfrac{3}{2}\left(ab+bc+ca\right)\Leftrightarrow ab+bc+ca\le\dfrac{3}{4}\)
từ (*)\(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2\sqrt{\dfrac{3}{4}+3}}=\dfrac{3}{\sqrt{5}}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{2}\)
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
Dự đoán dấu "=" khi \(a=b=c \Rightarrow P=28\)
Ta sẽ chứng minh \(P=28\) là GTNN
Thật vậy ta có: \(P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}-1+\dfrac{\left(a+b+c\right)^3}{abc}-27\ge0\)
\(\Leftrightarrow\dfrac{ab+bc+ca-\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\dfrac{\left(a+b+c\right)^3-27abc}{abc}\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3-27abc}{abc}-\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\dfrac{\dfrac{a+b+7c}{2}\cdot\left(a-b\right)^2}{abc}-\dfrac{\left(a-b\right)^2}{2\left(a^2+b^2+c^2\right)}\right)\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{a+b+7c}{2abc}-\dfrac{1}{2\left(a^2+b^2+c^2\right)}\right)\right)\ge0\) *Đúng*
Vậy ...
Lời giải:
Đặt \(\left(\frac{ab}{c}, \frac{bc}{a}, \frac{ca}{b}\right)=(x,y,z)\)
Khi đó: \(xy=b^2; yz=c^2; xz=a^2\). Bài toán trở về dạng:
Cho $x,y,z>0$ thỏa mãn: \(xy+yz+xz=1\)
Tìm GTNN của \(P=x+y+z\)
Thật vậy: Ta đã biết một BĐT quen thuộc theo AM-GM là:
\((x+y+z)^2\geq 3(xy+yz+xz)\)
\(\Rightarrow x+y+z\geq \sqrt{3(xy+yz+xz)}=\sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\)
Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Lời giải:
Theo BĐT Cauchy Schwarz:
\(ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}\)
\(\Rightarrow a+b+c\geq 3\)
Áp dụng BĐT AM-GM:
\(A=a-\frac{ca}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}\)
\(=(a+b+c)-\left(\frac{ac}{c+a^2}+\frac{ab}{a+b^2}+\frac{bc}{b+c^2}\right)\)
\(\geq (a+b+c)-\left(\frac{ac}{2a\sqrt{c}}+\frac{ab}{2b\sqrt{a}}+\frac{bc}{2c\sqrt{b}}\right)\)
\(A\geq (a+b+c)-\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\)
Cũng theo BĐT AM-GM:
\(\sqrt{a}+\sqrt{b}+\sqrt{c}\leq \frac{a+1}{2}+\frac{b+1}{2}+\frac{c+1}{2}=\frac{a+b+c+1}{4}\)
\(\Rightarrow A\geq a+b+c-\frac{a+b+c+3}{4}=\frac{3}{4}(a+b+c)-\frac{3}{4}\geq \frac{3}{4}.3-\frac{3}{4}=\frac{3}{2}\)
Vậy \(A_{\min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
Đề sai rồi: a,b,c > 0 thì làm sao mà có: ab + bc + ca = 0 được.
Ta có \(ab+bc+ca=2abc\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\\P=\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{y^3}{\left(2-y\right)^3}+\dfrac{z^3}{\left(2-z\right)^2}\end{matrix}\right.\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{x^3}{\left(2-x\right)^2}+\dfrac{2-x}{8}+\dfrac{2-x}{8}\ge3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(2-y\right)^2}+\dfrac{2-y}{8}+\dfrac{2-y}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(2-z\right)^2}+\dfrac{2-z}{8}+\dfrac{2-z}{8}\ge\dfrac{3z}{8}\end{matrix}\right.\)
\(\Rightarrow P+\dfrac{12-2\left(x+y+z\right)}{8}\ge\dfrac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{2}{3}\)