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\(\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}=\frac{1}{a^2+a^2+b^2}+\frac{1}{b^2+b^2+c^2}+\frac{1}{c^2+c^2+a^2}\)
\(< =\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\right)+\frac{1}{9}\left(\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{1}{9}\left(\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\)(bđt svacxo)
\(=\frac{1}{9}\left(\frac{1}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)=\frac{1}{9}\cdot3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
\(=\frac{1}{9}\cdot3\cdot\frac{1}{3}=\frac{1}{9}\cdot1=\frac{1}{9}\)
\(\Rightarrow\frac{1}{2a^2+b^2}+\frac{1}{2b^2+c^2}+\frac{1}{2c^2+a^2}< =\frac{1}{9}\)(đpcm)
dấu = xảy ra khi \(\frac{1}{a^2}=\frac{1}{b^2}=\frac{1}{c^2}=\frac{1}{9}\Rightarrow a=b=c=3\)
Đặt \(A=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)
Ta có : \(\frac{a}{b^2+c^2}=\frac{a}{3-a^2}=\frac{a}{\sqrt{\left(3-a^2\right)\left(3-a^2\right)}}=\frac{a^2}{a\sqrt{\left(3-a^2\right)\left(3-a^2\right)}}\)
\(=\frac{a^2\sqrt{2}}{\sqrt{2a^2\left(3-a^2\right)\left(3-a^2\right)}}\)
Theo BĐT Cô - si ta có :
\(0< \sqrt[3]{2a^2.\left(3-a^2\right).\left(3-a^2\right)}\le\frac{2a^2+3-a^2+3-a^2}{3}=2\)
\(\Leftrightarrow0< 2a^2.\left(3-a^2\right)\left(3-a^2\right)\le8\)
\(\Leftrightarrow0< \sqrt{2a^2\left(3-a^2\right)\left(3-a^2\right)}\le2\sqrt{2}\)
\(\Leftrightarrow\frac{a^2\sqrt{2}}{\sqrt{2a^2\left(3-a^2\right)\left(3-a^2\right)}}\ge\frac{a^2\sqrt{2}}{2\sqrt{2}}=\frac{a^2}{2}\)
Hay : \(\frac{a}{b^2+c^2}\ge\frac{a^2}{2}\)
Chứng minh tương tự ta có : \(\frac{b}{c^2+a^2}\ge\frac{b^2}{2};\frac{c}{a^2+b^2}\ge\frac{c^2}{2}\)
Do đó : \(A\ge\frac{1}{2}\left(a^2+b^2+c^2\right)=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vậy \(Min\) \(A=\frac{3}{2}\) khi \(a=b=c=1\)
Gọi biểu thức là N
Dự đoán \(MinN=\frac{3}{2}\)khi a = b = c = 1, ta dùng UCT giải quyết bài toán
Ta viết lại \(N=\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\)(do \(a^2+b^2+c^2=3\)theo giả thiết)
Xét bất đẳng thức phụ \(\frac{a}{3-a^2}\ge\frac{a^2}{2}\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{a\left(a+2\right)\left(a-1\right)^2}{2\left(3-a^2\right)}\ge0\)(Đúng vì \(3-a^2=b^2+c^2>0\)và a > 0)
Tương tự: \(\frac{b}{3-b^2}\ge\frac{b^2}{2}\)(1); \(\frac{c}{3-c^2}\ge\frac{c^2}{2}\)(2)
Cộng theo vế ba bất đẳng thức (*), (1) và (2), ta được: \(\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\ge\frac{a^2+b^2+c^2}{2}=\frac{3}{2}\)
Đẳng thức xảy ra khi a = b = c = 1
10. a)
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(x^4+y^4\right)=ab\left(x^2+y^2\right)^2\Leftrightarrow\left(bx^2-ay^2\right)^2=0\Leftrightarrow bx^2=ay^2\)
b) Từ \(ay^2=bx^2\Rightarrow\frac{y^2}{b}=\frac{x^2}{a}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\); \(\frac{y^{2008}}{b^{1004}}=\frac{1}{\left(a+b\right)^{1004}}\)
\(\Rightarrow\frac{x^{2008}}{a^{1004}}+\frac{y^{2008}}{b^{1004}}=\frac{2}{\left(a+b\right)^{1004}}\)
25. Ta có \(\left(ax+by+cz\right)^2=0\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(abxy+bcyz+acxz\right)\)
Xét mẫu số của P : \(bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)
\(=y^2bc-2bcyz+bcz^2+acx^2-2xzac+acz^2+abx^2-2abxy+aby^2\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(abxy+xzac+bcyz\right)\)
\(=y^2bc+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\)
\(=c\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+a\left(ax^2+by^2+cz^2\right)=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
\(\Rightarrow P=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{2007}\)
8. \(\frac{x^3}{a^3}+\frac{y^3}{b^3}=\left(\frac{x}{a}+\frac{y}{b}\right)^3-3.\frac{xy}{ab}\left(\frac{x}{a}+\frac{y}{b}\right)=1^3-3.\left(-2\right).1=7\)
Ta có:
y02 + ay0 + b = 0
\(\Leftrightarrow\)y04 = (ay0 + b)2
\(\le\)(a2 + b2)(y02 + 1)
\(\Rightarrow\)y04 - 1 < (a2 + b2)(y02 + 1)
\(\Rightarrow\)y02 - 1 < a2 + b2
\(\Rightarrow\)y02 < 1 + a2 + b2
3/ Dễ thấy \(0\le x,y,z\le1\)
Ta có:
x2 + y2 + z2 = x3 + y3 + z3
\(\Leftrightarrow\)x2(1 - x) + y2(1 - y) + z2(1 - z) = 0
Dấu = xảy ra khi (x, y, z) = (0,0,1) và các hoán vị của nó