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Chắc là a;b;c hết chứ?
\(VT=\dfrac{a}{a+b+c+b-a}+\dfrac{b}{a+b+c+c-b}+\dfrac{c}{a+b+c+a-c}\)
\(VT=\dfrac{a}{c+2b}+\dfrac{b}{a+2c}+\dfrac{c}{b+2a}=\dfrac{a^2}{ac+2ab}+\dfrac{b^2}{ab+2bc}+\dfrac{c^2}{bc+2ac}\)
\(VT\ge\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\) (đpcm)
cho x,y,z>0 ,x+y+z=1 chu nhi?
\(\Rightarrow\dfrac{x}{x+y+z+y-x}=\dfrac{x}{2y+z}\)
\(\Rightarrow\dfrac{y}{1+z-y}=\dfrac{y}{x+y+z+z-y}=\dfrac{y}{2z+x}\)
\(\Rightarrow\dfrac{z}{1+x-z}=\dfrac{z}{x+y+z+x-z}=\dfrac{z}{2x+y}\)
\(\Rightarrow A=\dfrac{x}{2y+z}+\dfrac{y}{2z+x}+\dfrac{z}{2x+y}=\dfrac{x^2}{2xy+xz}+\dfrac{y^2}{2zy+xy}+\dfrac{z^2}{2xz+xz}\ge\dfrac{\left(x+y+z\right)^2}{3\left(xy+yz+xz\right)}=1\)
dau"=" xay ra<=>x=y=z=1/3
\(a+\dfrac{1}{a+1}=\dfrac{a^2+a+1}{a+1}=\dfrac{4a^2+4a+4}{4\left(a+1\right)}=\dfrac{3\left(a+1\right)^2+\left(a-1\right)^2}{4\left(a+1\right)}\ge\dfrac{3\left(a+1\right)^2}{4\left(a+1\right)}=\dfrac{3}{4}\left(a+1\right)\ge\dfrac{3}{2}\sqrt{a}\)
Tương tự: \(b+\dfrac{1}{b+1}\ge\dfrac{3}{2}\sqrt{b}\) ; \(c+\dfrac{1}{c+1}\ge\dfrac{3}{2}\sqrt{c}\)
Nhân vế:
\(VT\ge\dfrac{27}{8}\sqrt{abc}\ge\dfrac{27}{8}\) (đpcm)
Áp dụng bđt AM - GM ta có :
\(\sqrt{b-1}\le\frac{b-1+1}{2}=\frac{b}{2}\Rightarrow a\sqrt{b-1}\le\frac{ab}{2}\)
\(\sqrt{a-1}\le\frac{a-1+1}{2}=\frac{a}{2}\Rightarrow b\sqrt{a-1}\le\frac{ba}{2}\)
\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)(đpcm)
b2 dễ tự lm
b2 x2 là x mũ 2. y2 là y mũ 2 .
yx−y=x2+2
yx−y−x2−2=0
x=−2−y+√y2−4y−8,−2−y−√y2−4y−8
x=−2−y+√y2−4y−8,−2−y−√y2−4y−8
x=−2−y+√y2−4y−8,−2−y−√y2−4y−8
k sau giúp tiếp
Let \(\left(a;b;c\right)\rightarrow\left(\frac{yz}{x^2};\frac{xz}{y^2};\frac{xy}{z^2}\right)\) we have:
\(\frac{x^4}{y^2z^2+x^2yz+x^4}+\frac{y^4}{x^2z^2+xy^2z+y^4}+\frac{z^4}{x^2y^2+xyz^2+z^4}\ge1\left(○\right)\)
By Cauchy-Schwarz: \(L-H-S_{\left(○\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{Σ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2}\)
Hence we need to prove: \(\frac{\left(x^2+y^2+z^2\right)^2}{Σ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2}\ge1\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2\geΣ_{cyc}x^4+Σ_{cyc}x^2yz+Σ_{cyc}y^2z^2\)
\(\Leftrightarrow x^2yz+xyz^2+xy^2z\ge x^2y^2+y^2z^2+z^2x^2\)
Follow AM-GM's ineq, it's enough to prove the last ineq
The equality occurs when \(a=b=c=1\)
Ta chứng minh: \(2\left(a^2-ab+b^2\right)^2\ge b^4+a^4\left(1\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)^2\ge0\)( Luôn đúng \(\forall a;b\))
Tương tự có: \(2\left(b^2-bc+c^2\right)^2\ge b^4+c^4\left(2\right)\)
Và: \(2\left(c^2-ca+a^2\right)^2\ge a^4+c^4\left(3\right)\)
Ta nhân các vế trên ta được: \(8\left(a^2-ab+b^2\right)^2\left(b^2-bc+c^2\right)^2\left(c^2-ca+a^2\right)^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)=8\)
Hay: \(\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\ge1\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Trâu bò:
Giả sử c = min{a,b,c}
Đặt a =x +c; b = y +c;c=c thì x,y >= 0
C/m: \(8\left[\left(a^2-ab+b^2\right)\left(b^2-bc+c^2\right)\left(c^2-ca+a^2\right)\right]^2\ge\left(a^4+b^4\right)\left(b^4+c^4\right)\left(c^4+a^4\right)\)
Xét hiệu hai vế thu được:
\(c*(12*x^3*y^8-8*x^4*y^7+16*x^5*y^6+16*x^6*y^5-8*x^7*y^4+12*x^8*y^3)+c^2*(18*x^2*y^8-16*x^3*y^7+60*x^4*y^6+60*x^6*y^4-16*x^7*y^3+18*x^8*y^2)+c^3*(12*x*y^8+16*x^2*y^7+88*x^4*y^5+88*x^5*y^4+16*x^7*y^2+12*x^8*y)+c^4*(6*y^8+16*x*y^7+32*x^2*y^6-32*x^3*y^5+242*x^4*y^4-32*x^5*y^3+32*x^6*y^2+16*x^7*y+6*x^8)+7*x^4*y^8+c^5*(16*y^7+16*x*y^6+88*x^3*y^4+88*x^4*y^3+16*x^6*y+16*x^7)-16*x^5*y^7+c^6*(24*y^6-16*x*y^5+60*x^2*y^4+60*x^4*y^2-16*x^5*y+24*x^6)+24*x^6*y^6+c^7*(16*y^5-8*x*y^4+16*x^2*y^3+16*x^3*y^2-8*x^4*y+16*x^5)-16*x^7*y^5+c^8*(8*y^4-16*x*y^3+24*x^2*y^2-16*x^3*y+8*x^4)+7*x^8*y^4\)Dấu " * " là nhân.
Dễ thấy nó đúng -> qed
Ko hieu het.Minh m jop 6 a
a=b=2
Dung 10000000000000000%