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\(P=\dfrac{a^2+b^2+c^2+2\left(ab+bc+ca\right)+1}{a+b+c-abc}=\dfrac{\left(a+b+c\right)^2+1}{a+b+c-abc}\ge\dfrac{\left(a+b+c\right)^2+1}{a+b+c}\)
\(\Rightarrow P\ge a+b+c+\dfrac{1}{a+b+c}\) (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
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\(\left(a^2+b^2+c^2\right)^2\ge a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2\)
\(\ge a^4+b^4+c^4+a^2b^2-2abc^2\)
\(=\left(a^2+b^2+c^2\right)\left(a^4+b^4+\left(c^2-ab\right)^2\right)\)
\(\ge\left(a^3+b^3+c\left(c^2-ab\right)\right)^2\)
\(=\left(a^3+b^3+c^3-abc\right)^2\ge\left(a^3+b^3+c^3-3abc\right)^2=1\)
\(\Rightarrow B\ge1\)
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Áp dụng BĐT Am-GM \(\frac{a^3+b^3+c^3}{a^2+b^2+c^2}\ge\frac{3\sqrt[3]{\left(abc\right)^3}}{3\sqrt[3]{\left(abc\right)^2}}=2\)
dấu = xảy ra khi a=b=c=2
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Bài toán số 41 có 2 cách làm, tôi làm cách thứ 2
Đặt \(Q=\sqrt{\frac{x}{y+z}}+\sqrt{\frac{y}{x+z}}+\sqrt{\frac{z}{x+y}}\)\(\Rightarrow Q^2=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}+2\left(\sqrt{\frac{xy}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{yz}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\right)\)ta thấy rằng \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{1}{4}\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\left(xy+yz+zx\right)\)
\(=\frac{x^2+y^2+z^2}{4}+\frac{xyz}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{x^2+y^2+z^2}{4}\)
Áp dụng bất đẳng thức AM-GM ta có \(\sqrt{\frac{yx}{\left(z+x\right)\left(x+y\right)}}\ge\frac{2yx}{2\sqrt{\left(xy+yz\right)\left(yz+yx\right)}}\ge\frac{2xy}{2xy+yz+xz}\ge\frac{2xy}{2\left(xy+yz+zx\right)}=\frac{xy}{xy+yz+zx}\)
Tương tự ta có \(\hept{\begin{cases}\sqrt{\frac{yz}{\left(z+x\right)\left(z+y\right)}}\ge\frac{yz}{xy+yz+zx}\\\sqrt{\frac{xz}{\left(x+y\right)\left(y+z\right)}}\ge\frac{xz}{xy+yz+zx}\end{cases}}\)
\(\Rightarrow\sqrt{\frac{xy}{\left(y+z\right)\left(z+x\right)}}+\sqrt{\frac{yz}{\left(z+x\right)\left(x+y\right)}}+\sqrt{\frac{zx}{\left(x+y\right)\left(y+z\right)}}\ge1\)nên \(Q\ge\sqrt{\frac{x^2+y^2+z^2}{4}+2}\)
\(\Rightarrow Q\ge\sqrt{\frac{x^2+y^2+z^2}{2}+4}+\frac{4}{\sqrt{x^2+y^2+z^2}}\)
Đặt \(t=\sqrt{x^2+y^2+z^2}\Rightarrow t\ge\sqrt{xy+yz+zx}=2\)
Xét hàm số g(t)=\(\sqrt{\frac{t^2}{2}+4}+\frac{4}{t}\left(t\ge2\right)\)khi đó ta có
\(g'\left(t\right)=\frac{t}{2\sqrt{\frac{t^2}{2}+4}}-\frac{4}{t^2};g'\left(t\right)=0\Leftrightarrow t^6-32t^2-256=0\Leftrightarrow t=2\sqrt{2}\)
Lập bảng biến thiên ta có min[2;\(+\infty\)) \(g\left(t\right)=g\left(2\sqrt{2}\right)=3\sqrt{2}\)
Hay minS=\(3\sqrt{2}\)<=> a=c=1; b=2
Đặt a=xc; b=cy (x;y >=1)
- Thay x=1 vào giả thiết ta có \(\sqrt{b-c}=\sqrt{b}\Rightarrow c=0\) (không thỏa mãn vì c>0)
- Thay y=1 vào giả thiết ta có \(\sqrt{a-c}=\sqrt{a}\Rightarrow c=0\)( không thỏa mãn vì c>0)
- Xét x,y>1 thay vào giả thiết ta có
\(\sqrt{x-1}+\sqrt{y-1}=\sqrt{xy}\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=xy\)
\(\Leftrightarrow xy-x-y+1-2\sqrt{\left(x-1\right)\left(y-1\right)}+1=0\)
\(\Leftrightarrow\left(\sqrt{\left(x-1\right)\left(y-1\right)}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=1\Leftrightarrow xy=x+y\ge2\sqrt{xy}\Rightarrow xy\ge4\)
Biểu thức P được viết lại như sau
\(P=\frac{x}{y+1}+\frac{y}{x+1}+\frac{1}{x+y}+\frac{1}{x^2+y^2}=\frac{x^2}{xy+x}+\frac{y^2}{xy+y}+\frac{1}{x^2+y^2}+\frac{1}{\left(x+y\right)^2-2xy}\)
\(P\ge\frac{\left(x+y\right)^2}{2xy+x+y}+\frac{1}{x+y}+\frac{1}{\left(x+y\right)^2-2xy}=\frac{xy}{3}+\frac{1}{xy}+\frac{1}{x^2y^2-2xy}=\frac{x^3y^3-2x^2y^2+3xy-3}{3\left(x^2y^2-2xy\right)}\)
Đặt t=xy với t>=4
Xét hàm số \(f\left(t\right)=\frac{t^3-2t^2+3t-3}{t^2-2t}\left(t\ge4\right)\)
Ta có \(f'\left(t\right)=\frac{t^4-4t^3+t^2+6t-6}{\left(t^2-2t\right)^2}=\frac{t^3\left(t-4\right)+6\left(t-4\right)+18}{\left(t^2-2t\right)^2}>0\forall t\ge4\)
Lập bảng biến thiên ta có \(minf\left(t\right)=f\left(4\right)=\frac{41}{8}\)
Vậy \(minP=\frac{41}{24}\)khi x=y=z=2 hay a=b=2c