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Biểu thức này có vẻ chỉ tìm được min chứ ko tìm được max:
Min:
\(P^2=a+b+c+a^3b^3+b^3c^3+c^3a^3+2\sqrt{\left(a+b^3c^3\right)\left(b+c^3a^3\right)}+2\sqrt{\left(a+b^3c^3\right)\left(c+a^3b^3\right)}+2\sqrt{\left(b+c^3a^3\right)\left(c+a^3b^3\right)}\)
\(P^2\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\ge a+b+c=2\)
\(\Rightarrow P\ge\sqrt{2}\)
\(P_{min}=\sqrt{2}\) khi \(\left(a;b;c\right)=\left(0;0;2\right)\) và các hoán vị
a2+b2+c2=4−abc≤4
Smax=4 khi 1 trong 3 số bằng 0
4=abc+a2+b2+c2≥abc+33√(abc)2
Đặt 3√abc=x>0⇒x3+3x2−4≤0
⇔(x−1)(x+2)2≤0⇒x≤1
⇒abc≤1⇒S=4−abc≥3
Dấu "=" xảy ra khi a=b=c=1
Min là hoán vị a=b=0 c=2 ; a=c=0 b=2 ; b=c=0 a=2 mà :vv
mà thôi Min làm đr còn max
TKS
Áp dụng Bất đẳng thức Cauchy cho 3 số thực dương ta có :
\(a^2b+b^2c+c^2a\ge3\sqrt[3]{a^2bb^2cc^2a}=3\sqrt[3]{a^3b^3c^3}=3abc\)
Khi đó :\(P\ge3abc=\left(a+b+c\right)\left(abc\right)\)
...
Min:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\ge a^3+b^3+c^3\)
\(\Rightarrow a+b+c\ge\sqrt[3]{a^3+b^3+c^3}=\sqrt[3]{3}\)
\(\Rightarrow P=\dfrac{a}{7-3bc}+\dfrac{b}{7-3ca}+\dfrac{c}{7-3ab}\ge\dfrac{a}{7}+\dfrac{b}{7}+\dfrac{c}{7}=\dfrac{a+b+c}{7}\ge\dfrac{\sqrt[3]{3}}{7}\)
Dấu "=" xảy ra tại \(\left(a;b;c\right)=\left(0;0;\sqrt[3]{3}\right)\) và các hoán vị
Max:
\(\left(a^3+1+1\right)+\left(b^3+1+1\right)+\left(c^3+1+1\right)\ge3a+3b+3c\)
\(\Rightarrow a+b+c\le\dfrac{a^3+b^3+c^3+6}{3}=3\)
Khi đó:
\(7P=\dfrac{7a}{7-3bc}+\dfrac{7b}{7-3ca}+\dfrac{7c}{7-3ab}=\dfrac{a\left(7-3bc\right)+3abc}{7-3bc}+\dfrac{b\left(7-3ca\right)+3abc}{7-3ca}+\dfrac{c\left(7-3ab\right)+3abc}{7-3ab}\)
\(=a+b+c+\dfrac{3abc}{7-3bc}+\dfrac{3abc}{7-3ca}+\dfrac{3abc}{7-3ab}\)
Ta có:
\(7-3ab\ge\dfrac{7}{9}\left(a+b+c\right)^2-3ab=\dfrac{1}{9}\left[\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)+7c^2+14bc+14ca\right]\)
Do \(\dfrac{13}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(a^2+b^2\right)\ge\dfrac{1}{2}\left(a^2+b^2\right)\ge ab\)
\(\Rightarrow7-3ab\ge\dfrac{1}{9}\left(ab+7c^2+14bc+14ca\right)\)
\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{27abc}{ab+7c\left(c+2a+2b\right)}\le\dfrac{27abc}{36^2}\left(\dfrac{1^2}{ab}+\dfrac{35^2}{7c\left(c+2a+2b\right)}\right)\)
\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{c+2a+2b}=\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{\left(a+b+c\right)+\left(a+b\right)}\)
\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{175}{48}.\dfrac{ab}{5^2}\left(\dfrac{3^2}{a+b+c}+\dfrac{2^2}{a+b}\right)\)
\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{12}.\dfrac{ab}{a+b}\le\dfrac{c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}+\dfrac{7}{48}.\dfrac{\left(a+b\right)^2}{a+b}\)
\(\Rightarrow\dfrac{3abc}{7-3ab}\le\dfrac{7a+7b+c}{48}+\dfrac{21}{16}.\dfrac{ab}{a+b+c}\)
Tương tự:
\(\dfrac{3abc}{7-3bc}\le\dfrac{a+7b+7c}{48}+\dfrac{21}{16}.\dfrac{bc}{a+b+c}\)
\(\dfrac{3abc}{7-3ca}\le\dfrac{7a+b+7c}{48}+\dfrac{21}{16}.\dfrac{ca}{a+b+c}\)
\(\Rightarrow7P\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{16}\left(\dfrac{ab+bc+ca}{a+b+c}\right)\le\dfrac{21}{16}\left(a+b+c\right)+\dfrac{21}{48}.\dfrac{\left(a+b+c\right)^2}{a+b+c}\)
\(\Rightarrow7P\le\dfrac{7}{4}\left(a+b+c\right)\)
\(\Rightarrow P\le\dfrac{a+b+c}{4}\le\dfrac{3}{4}\)
Vậy \(P_{max}=\dfrac{3}{4}\) khi \(a=b=c=1\)
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
1,\(T=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=20\left(a^2-ab+b^2\right)=\)
\(=10\left(a^2-2ab+b^2\right)+10\left(a^2+b^2\right)\)
\(\ge10\left(a-b\right)^2+5.\left(a+b\right)^2\ge0+5.20^2=2000\)
2,a,\(\sqrt{a}+\sqrt{b-1}+\sqrt{c-2}=\frac{1}{2}\left(a+b+c\right)\)
\(\Leftrightarrow a-2\sqrt{a}+b-2\sqrt{b-1}+c-2\sqrt{c-2}=0\)
\(\Leftrightarrow a-2\sqrt{a}+1+b-1-2\sqrt{b-1}+1+c-2+2\sqrt{c-2}+1=0\)
\(\Leftrightarrow\left(\sqrt{a}-1\right)^2+\left(\sqrt{b-1}-1\right)^2+\left(\sqrt{c-2}-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)
b,sai đề
Xét \(\frac{a+b}{2}\ge\sqrt{ab}\Rightarrow10\ge\sqrt{ab}\Leftrightarrow100\ge ab\)
\(T=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=20\left(a^2-ab+b^2\right)=20\left[a^2+2ab+b^2-3ab\right]=20\left(20\right)^2-6ab\)
\(T\ge20.20^2-6.100=7400\)
Lời giải:
Tìm min:
Áp dụng BĐT AM-GM:
$a^3+a^3+1\geq 3a^2$
$b^3+b^3+1\geq 3b^2$
$c^3+c^3+1\geq 3c^2$
$\Rightarrow 2(a^3+b^3+c^3)+3\geq 3(a^2+b^2+c^2)$
$\Leftrightarrow 2P+3\geq 9$
$\Leftrightarrow P\geq 3$
Vậy $P_{\min}=3$ khi $(a,b,c)=(1,1,1)$
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Tìm max:
$a^2+b^2+c^2=3\Rightarrow a^2,b^2,c^2\leq 3$
$\Rightarrow a,b,c\leq \sqrt{3}$
Do đó: $a^3-\sqrt{3}a^2=a^2(a-\sqrt{3})\leq 0$
$\Rightarrow a^3\leq \sqrt{3}a^2$
Tương tự với $b,c$ và cộng theo vế:
$P\leq \sqrt{3}(a^2+b^2+c^2)=3\sqrt{3}$
Vậy $P_{\max}=3\sqrt{3}$ khi $(a,b,c)=(\sqrt{3},0,0)$ và hoán vị.
\(a^3+a^3+1\ge3\sqrt[3]{a^3.a^3.1}=3a^2\)
Tương tự: \(2b^3+1\ge3b^2\) ; \(2c^3+1\ge3c^2\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge3\left(a^2+b^2+c^2\right)=9\)
\(\Rightarrow a^3+b^3+c^3\ge3\)
\(A_{min}=3\) khi \(a=b=c=1\)
Lại có: \(\left\{{}\begin{matrix}a;b;c\ge0\\a^2+b^2+c^2=3\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le\sqrt{3}\)
\(\Rightarrow a^2\left(a-\sqrt{3}\right)\le0\Rightarrow a^3\le\sqrt{3}a^2\)
Tương tự: \(b^3\le\sqrt{3}b^2\) ; \(c^3\le\sqrt{3}c^2\)
\(\Rightarrow a^3+b^3+c^3\le\sqrt{3}\left(a^2+b^2+c^2\right)=3\sqrt{3}\)
\(A_{max}=3\sqrt{3}\) khi \(\left(a;b;c\right)=\left(0;0;\sqrt{3}\right)\) và các hoán vị