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\(a^3+a^3+1\ge3\sqrt[3]{a^3.a^3.1}=3a^2\)
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
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
\(\left(a^2+b^2+c^2\right)^2\ge a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2\)
\(\ge a^4+b^4+c^4+a^2b^2-2abc^2\)
\(=\left(a^2+b^2+c^2\right)\left(a^4+b^4+\left(c^2-ab\right)^2\right)\)
\(\ge\left(a^3+b^3+c\left(c^2-ab\right)\right)^2\)
\(=\left(a^3+b^3+c^3-abc\right)^2\ge\left(a^3+b^3+c^3-3abc\right)^2=1\)
\(\Rightarrow B\ge1\)
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.
Ta sẽ chứng minh BĐT sau: \(\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\ge\dfrac{10}{\left(a+b+c\right)^2}\)
Do vai trò a;b;c như nhau, ko mất tính tổng quát, giả sử \(c=min\left\{a;b;c\right\}\)
Đặt \(\left\{{}\begin{matrix}x=a+\dfrac{c}{2}\\y=b+\dfrac{c}{2}\end{matrix}\right.\) \(\Rightarrow x+y=a+b+c\)
Đồng thời \(b^2+c^2=\left(b+\dfrac{c}{2}\right)^2+\dfrac{c\left(3c-4b\right)}{4}\le\left(b+\dfrac{c}{2}\right)^2=y^2\)
Tương tự: \(a^2+c^2\le x^2\) ; \(a^2+b^2\le x^2+y^2\)
Do đó: \(A\ge\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{10}{\left(x+y\right)^2}\)
Mà \(\dfrac{1}{\left(x+y\right)^2}\le\dfrac{1}{4xy}\) nên ta chỉ cần chứng minh:
\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\ge\dfrac{5}{2xy}\)
\(\Leftrightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{2}{xy}+\dfrac{1}{x^2+y^2}-\dfrac{1}{2xy}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-y\right)^2}{x^2y^2}-\dfrac{\left(x-y\right)^2}{2xy\left(x^2+y^2\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(2x^2+2y^2-xy\right)}{2x^2y^2}\ge0\) (luôn đúng)
Vậy \(A\ge\dfrac{10}{\left(a+b+c\right)^2}\ge\dfrac{10}{3^2}=\dfrac{10}{9}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};\dfrac{3}{2};0\right)\) và các hoán vị của chúng
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
