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\(\left(1+a^3\right)\left(1+b^3\right)\left(1+b^3\right)\ge\left(1+ab^2\right)^3\)
\(\Leftrightarrow\)\(\frac{1+a^3}{1+ab^2}\ge\frac{\left(1+ab^2\right)^2}{\left(1+b^3\right)^2}\)
\(\Rightarrow\)\(3P\ge\Sigma\frac{\left(1+ab^2\right)^2}{\left(1+b^3\right)^2}+2\Sigma\frac{1+a^3}{1+ab^2}\ge9\sqrt[9]{\frac{\Pi\left(1+ab^2\right)^2}{\Pi\left(1+a^3\right)^2}\left(\frac{\Pi\left(1+a^3\right)}{\Pi\left(1+ab^2\right)}\right)^2}=9\)
\(\Rightarrow\)\(P\ge3\)
dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT=A+B và xét
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\text{∑}\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\text{∑}\left(3a-\frac{3ab}{2}\right)\)
\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\text{∑}\left(1-\frac{b^2}{1+b^2}\right)\ge\text{∑}\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\text{∑}ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)
(Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}=3\))
Dấu = khi a=b=c=1
\(VT=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}\)
Ta tách VT = A + b và xét :
\(A=\frac{3a}{1+b^2}+\frac{3b}{1+c^2}+\frac{3c}{1+a^2}=\Sigma\left(3a-\frac{3ab^2}{1+b^2}\right)\ge\Sigma\left(3a-\frac{3ab}{2}\right)\)\(B=\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+a^2}=\Sigma\left(1-\frac{b^2}{1+b^2}\right)\ge\Sigma\left(1-\frac{b}{2}\right)\)
\(\Rightarrow VT=A+B=3+\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\Sigma ab=\frac{5}{2}\left(a+b+c\right)-\frac{3}{2}\ge\frac{15}{2}-\frac{3}{2}=6\)( Do \(a+b+c\ge\sqrt{3\left(ab+bc+ca\right)=3}\))
Dấu = khi a = b = c = 1 .
Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)
\(\ge\left(1+3a\right)\left(1-\frac{b^2}{2b}\right)=\left(1+3a\right)\left(1-\frac{b}{2}\right)\)
\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)
Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)
\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)
Đẳng thức xảy ra khi a = b = c = 1
Áp dụng Bđt \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có:
\(P\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ca}\)
Lại có:
\(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}\)
\(\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=9\)
Mặt khác \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\)
\(\Rightarrow\frac{1}{ab+bc+ca}\ge3\)\(\Rightarrow P_{Min}=30\)
Dấu = khi \(a=b=c=\frac{1}{3}\)
Từ giả thiết và BĐT AM-GM suy ra:\(\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)\(\ge\)3
Ta có:
P\(\ge\)\(\frac{2a^3}{3\left(a^2+b^2\right)}\)+\(\frac{2b^3}{3\left(c^2+b^2\right)}\)+\(\frac{2c^3}{3\left(a^2+c^2\right)}\)
=\(\frac{2}{3}\)(\(\frac{a\left(a^2+b^2\right)-ab^2}{\left(a^2+b^2\right)}\)+\(\frac{b\left(c^2+b^2\right)-bc^2}{\left(c^2+b^2\right)}\)+\(\frac{a\left(a^2+c^2\right)-ca^2}{\left(a^2+c^2\right)}\))
=\(\frac{2}{3}\)(a+b+c-\(\frac{ab^2}{\left(a^2+b^2\right)}\)-\(\frac{bc^2}{\left(c^2+b^2\right)}\)-\(\frac{ca^2}{\left(a^2+c^2\right)}\))
\(\ge\)\(\frac{2}{3}\)(a+b+c-\(\frac{a}{2}\)-\(\frac{b}{2}\)-\(\frac{c}{2}\))
=\(\frac{2}{3}\).\(\frac{a+b+c}{2}\)=\(\frac{a+b+c}{3}\)=\(\frac{\left(a+1\right)+\left(b+1\right)+\left(c+1\right)}{3}\)-1
\(\ge\)\(\frac{3\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}{3}\)-1\(\ge\)2
Vậy:MinP=2 khi a=b=c=2
cách này dễ hiểu hơn nè :
Áp dụng BĐT : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
Ta có : \(1\ge\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)
\(\Leftrightarrow1\ge\frac{9}{a+b+c+3}\)\(\Leftrightarrow a+b+c+3\ge9\)\(\Leftrightarrow a+b+c\ge6\)
\(\frac{a^3}{a^2+ab+b^2}=\frac{a\left(a^2+ab+b^2\right)-ab^2-a^2b}{a^2+ab+b^2}=a-\frac{ab^2+a^2b}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3ab}=a-\frac{a+b}{3}\)
Tương tự : \(\frac{b^3}{b^2+bc+c^2}\ge b-\frac{b+c}{3}\); \(\frac{c^3}{c^2+ac+a^2}\ge c-\frac{a+c}{3}\)
Cộng cả 3 vế , ta được : \(P\ge a+b+c-\frac{2\left(a+b+c\right)}{3}=\frac{1}{3}\left(a+b+c\right)\ge\frac{1}{3}.6=2\)
Vậy GTNN của P là 2 \(\Leftrightarrow a=b=c=2\)
\(P=\frac{a^3+b^3+c^3}{2abc}+\frac{a^2c+b^2c}{c^3+abc}+\frac{b^2a+c^2a}{a^3+abc}+\frac{c^2b+a^2b}{b^3+abc}\)
\(\ge\frac{a^3}{2abc}+\frac{b^3}{2abc}+\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}+\frac{2abc}{a^3+abc}+\frac{2abc}{b^3+abc}\)
\(=\left(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}\right)+\left(\frac{b^3}{2abc}+\frac{2abc}{b^3+abc}\right)+\left(\frac{c^3}{2abc}+\frac{2abc}{c^3+abc}\right)\)
Xét: \(\frac{a^3}{2abc}+\frac{2abc}{a^3+abc}=\frac{a^3}{2abc}+\frac{1}{2}+\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}-\frac{1}{2}\ge2\sqrt{\left(\frac{a^3}{2abc}+\frac{1}{2}\right).\frac{1}{\frac{a^3}{2abc}+\frac{1}{2}}}-\frac{1}{2}=\frac{3}{2}\)
Tương tự với 2 cặp còn lại
Vậy ta có: \(P\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}=\frac{9}{2}\)
"=" xảy ra <=> a=b=c
Lời giải:
$A=a-\frac{ac}{c+a^2}+b-\frac{ab}{a+b^2}+c-\frac{bc}{b+c^2}$
$=\sum a-\sum \frac{ac}{c+a^2}$
Áp dụng BĐT AM-GM: $c+a^2\geq 2a\sqrt{c}$
$\Rightarrow A\geq \sum a-\frac{1}{2}\sum \sqrt{c}$
Áp dụng BĐT Cauchy-Schwarz:
$(\sum \sqrt{c})^2\leq (c+a+b)(1+1+1)$
$\Rightarrow \sum \sqrt{c}\leq 3\sum a$
Do đó $A\geq \sum a-\frac{1}{2}\sqrt{3\sum a}$
Đặt $\sqrt{3\sum a}=t$ thì $A\geq \frac{t^2}{3}-\frac{t}{2}(*)$
Từ điều kiện $ab+bc+ac=3abc\Rightarrow 3=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Áp dụng BĐT Cauchy-Schwarz:
$3=\sum \frac{1}{a}\geq \frac{9}{\sum a}\Rightarrow \sum a\geq 3$
$\Rightarrow t=\sqrt{3\sum a}\geq 3$
Do đó:
$\frac{t^2}{3}-\frac{t}{2}=(t-3)(\frac{t}{3}+\frac{1}{2})+\frac{3}{2}\geq \frac{3}{2}$ với mọi $t\geq 3(**)$
Từ $(*); (**)\Rightarrow A\geq \frac{3}{2}$
Vậy $A_{\min}=\frac{3}{2}$ khi $a=b=c=1$
Gọi \(S=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+ab+c^2}+\frac{a^3}{c^2+ab+a^2}\)
Dễ thấy \(P-S=0\)
\(\Rightarrow2P=\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{b^2+ab+c^2}+\frac{c^3+a^3}{c^2+ab+a^2}\)
Ta chứng minh:
\(\frac{a^3+b^3}{a^2+ab+b^2}\ge\frac{a+b}{3}\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\)(đúng)
\(\Rightarrow2P\ge\frac{a+b}{3}+\frac{b+c}{3}+\frac{c+a}{3}=\frac{2\left(a+b+c\right)}{3}=2\)
\(\Rightarrow P\ge1\)
Theo em nghĩ bài này ko thiếu điều kiện đâu cô quản lí ạ !!!
Áp dụng BĐT Bunhiacopxki ta có:
\(\left(ab+1\right)^2\le\left(a^2+1\right)\left(b^2+1\right)\)
Áp dụng BĐT AM-GM, ta có:
\(a^2+1=a.a.1+1\le\frac{a^3+a^3+1}{3}+1=\frac{2.\left(a^3+2\right)}{3}\)
\(b^2+1=b.b.1+1\le\frac{b^3+b^3+1}{3}+1=\frac{2.\left(b^3+2\right)}{3}\)
Do đó:
\(\left(ab+1\right)^2\le\frac{4}{9}\left(a^3+2\right)\left(b^3+2\right)\)
\(\Rightarrow ab+1\le\frac{2}{3}\sqrt{\left(a^3+2\right)\left(b^3+2\right)}\)
\(\Rightarrow\frac{a^3+2}{ab+1}\ge\frac{3}{2}.\sqrt{\frac{a^3+2}{b^3+2}}\) \(\left(1\right)\)
Tương tự, ta có:
\(\frac{b^3+2}{bc+1}\ge\frac{3}{2}.\sqrt{\frac{b^3+2}{c^3+2}}\) \(\left(2\right)\)
\(\frac{c^3+2}{ca+1}\ge\frac{3}{2}.\sqrt{\frac{c^3+2}{a^3+2}}\) \(\left(3\right)\)
Cộng theo vế của \(\left(1\right)\), \(\left(2\right)\) và \(\left(3\right)\) và áp dụng BĐT AM-GM, ta có:
\(G\ge\frac{3}{2}\left(\sqrt{\frac{a^3+2}{b^3+2}}+\sqrt{\frac{b^3+2}{c^3+2}}+\sqrt{\frac{c^3+2}{a^3+2}}\right)\) \(\ge\frac{3}{2}.3\sqrt[3]{\sqrt{\frac{a^3+2}{b^3+2}}.\sqrt{\frac{b^3+2}{c^3+2}}.\sqrt{\frac{c^3+2}{a^3+2}}}=\frac{9}{2}\)
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)
Vậy: \(G_{min}=\frac{9}{2}\Leftrightarrow a=b=c=1\)
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