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\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\right)\)
\(P\ge\left(a+b+c\right)^2\left(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\right)\)
\(P\ge\left(a+b+c\right)^2\left(\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{\dfrac{1}{3}\left(a+b+c\right)^2}\right)=30\)
\(P_{min}=30\) khi \(a=b=c\)
\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)
\(P\ge\dfrac{3}{2}+2\left(\dfrac{a^2+b^2}{a^2+c^2+b^2+c^2}+\dfrac{b^2+c^2}{a^2+b^2+a^2+c^2}+\dfrac{a^2+c^2}{a^2+b^2+b^2+c^2}\right)\)
Đặt \(\left(a^2+b^2;b^2+c^2;a^2+c^2\right)=\left(x;y;z\right)\)
\(\Rightarrow P\ge\dfrac{3}{2}+2\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{3}{2}+2\left(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{yz+xy}+\dfrac{z^2}{xz+yz}\right)\)
\(P\ge\dfrac{3}{2}+\dfrac{2\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3}{2}+\dfrac{3\left(xy+yz+zx\right)}{xy+yz+zx}=3+\dfrac{3}{2}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(a=b=c\)
Ta có: \(\sqrt{a^2+b^2+c^2}\ge\sqrt{\dfrac{\left(a+b+c\right)^2}{3}}=\sqrt{3};\sqrt{a^2+b^2+c^2}\le\sqrt{\left(a+b+c\right)^2}=3\).
Đặt \(\sqrt{a^2+b^2+c^2}=t\) \((\sqrt{3}\leq t\leq 3)\).
Ta có: \(P=t+\dfrac{9-t^2}{4}+\dfrac{1}{t^2}=\dfrac{4t^3+9t^2-t^4+4}{4t^2}\).
\(\Rightarrow P-\dfrac{28}{9}=\dfrac{\left(3-t\right)\left(9t^3-9t^2+4t+12\right)}{36}\).
Do \(\sqrt{3}\le t\le3\) nên \(3-t\geq 0\); \(9t^3-9t^2+4t+12>4t+12>0\).
Nên \(P\ge\dfrac{28}{9}\).
Đẳng thức xảy ra khi t = 3, tức (a, b, c) = (0; 0; 3) và các hoán vị.
Vậy...
\(P=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)+1}{a+b+c-abc}=\dfrac{\left(a+b+c\right)^2+1}{a+b+c-abc}\ge\dfrac{\left(a+b+c\right)^2+1}{a+b+c}\)
\(\Rightarrow P\ge a+b+c+\dfrac{1}{a+b+c}\) (1)
\(P=\dfrac{a^2+b^2+c^2+3\left(ab+bc+ca\right)}{\left(a+b+c\right)\left(ab+bc+ca\right)-abc}=\dfrac{\left(a+b\right)\left(b+c\right)+\left(b+c\right)\left(c+a\right)+\left(a+b\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(P=\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{a+b+c}\left(\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+c+b}{a+c}\right)\)
\(P=\dfrac{1}{a+b+c}\left(3+\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\dfrac{1}{a+b+c}\left(3+\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\right)\)
\(P\ge\dfrac{1}{a+b+c}\left(3+\dfrac{\left(a+b+c\right)^2}{2}\right)=\dfrac{3}{a+b+c}+\dfrac{a+b+c}{2}\)
\(\Rightarrow3P\ge\dfrac{3}{2}\left(a+b+c\right)+\dfrac{9}{a+b+c}\) (2)
Cộng vế (1) và (2):
\(\Rightarrow4P\ge\dfrac{5}{2}\left(a+b+c\right)+\dfrac{10}{a+b+c}\ge2\sqrt{\dfrac{50\left(a+b+c\right)}{2\left(a+b+c\right)}}=10\)
\(\Rightarrow P\ge\dfrac{5}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;1;0\right)\) và các hoán vị
Lời giải:Áp dụng BĐT AM-GM và BĐT Cauchy-Schwarz:
\(\frac{bc}{a^2+1}=\frac{bc}{(a^2+b^2)+(a^2+c^2)}\leq \frac{1}{4}.\frac{(b+c)^2}{(a^2+b^2)+(a^2+c^2)}\leq \frac{1}{4}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)\)
Hoàn toàn tương tự với các phân thức còn lại, ta có:
\(P\leq \frac{1}{4}\left(\frac{b^2+a^2}{a^2+b^2}+\frac{c^2+a^2}{a^2+c^2}+\frac{b^2+c^2}{b^2+c^2}\right)=\frac{3}{4}\)
(đpcm)
Dấu "=" xảy ra khi $a=b=c=\sqrt{\frac{1}{3}}$
Đẳng thức quen thuộc: \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\) và tương tự cho các mẫu số còn lại
Ta có:
\(\sum\dfrac{1}{a^2+1}=\sum\dfrac{1}{\left(a+b\right)\left(a+c\right)}=\dfrac{2\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2\left(ab+bc+ca\right)\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Mặt khác:
\(2\left(ab+bc+ca\right)\left(a+b+c\right)=\left[a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)\right]\left(a+b+c\right)\)
\(\ge\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2\) (Bunhiacopxki)
\(\Rightarrow\sum\dfrac{1}{a^2+1}\ge\dfrac{\left(a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\left(\dfrac{a}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\right)^2\)
\(=\left(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\right)^2\)
Do đó ta chỉ cần chứng minh:
\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{3}{2}\)
Đúng theo AM-GM:
\(\sum\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
\(P=\dfrac{a^3}{b^2+ab+bc+ca}+\dfrac{b^3}{c^2+ab+bc+ca}+\dfrac{c^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}\)
Ta có:
\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge\dfrac{3a}{4}\)
\(\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3b}{4}\)
\(\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge\dfrac{3c}{4}\)
Cộng vế:
\(P+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)
\(\Rightarrow P\ge\dfrac{1}{4}\left(a+b+c\right)\ge\dfrac{1}{4}.\sqrt{3\left(ab+bc+ca\right)}=\dfrac{\sqrt{3}}{4}\)
giúp em giải dấu "=" với ạ :<