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1.
Ta có:
\(x^4+y^4\ge\dfrac{1}{2}\left(x^2+y^2\right)^2=\dfrac{1}{2}\left(x^2+y^2\right)\left(x^2+y^2\right)\ge\left(x^2+y^2\right)xy\)
Đặt vế trái của BĐT cần chứng minh là P, áp dụng bồ đề vừa chứng minh ta có:
\(P\le\dfrac{a.abc}{bc\left(b^2+c^2\right)+a.abc}+\dfrac{b.abc}{ca\left(c^2+a^2\right)+b.abc}+\dfrac{c.abc}{ab\left(a^2+b^2\right)+c.abc}\)
\(P\le\dfrac{a^2.bc}{bc\left(a^2+b^2+c^2\right)}+\dfrac{b^2.ac}{ca\left(a^2+b^2+c^2\right)}+\dfrac{c^2.ab}{ab\left(a^2+b^2+c^2\right)}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
2.
\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}=1\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2}{3}\)
Bài 2:
a) Áp dụng BĐT AM - GM ta có:
\(\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{1}{4a}+\dfrac{1}{4b}\) \(\ge2\sqrt{\dfrac{1}{4^2ab}}=\dfrac{2}{4\sqrt{ab}}=\dfrac{1}{2\sqrt{ab}}\)
\(\ge\dfrac{1}{a+b}\) (Đpcm)
b) Trừ 1 vào từng vế của BĐT ta được BĐT tương đương:
\(\left(\frac{x}{2x+y+z}-1\right)+\left(\frac{y}{x+2y+z}-1\right)+\left(\frac{z}{x+y+2z}-1\right)\le\frac{-9}{4}\)
\(\Leftrightarrow-\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\le-\frac{9}{4}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)
Áp dụng BĐT phụ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\)
\(\ge\dfrac{9}{2x+y+z+x+2y+z+x+y+2z}=\dfrac{9}{4\left(x+y+z\right)}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\ge\frac{9}{4}\)
\(\Leftrightarrow\dfrac{x}{2x+y+z}+\dfrac{y}{x+2y+z}+\dfrac{z}{x+y+2z}\le\dfrac{3}{4}\) (Đpcm)
Bài 1:
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT\ge\dfrac{\left(a+b\right)^2}{a-1+b-1}=\dfrac{\left(a+b\right)^2}{a+b-2}\)
Nên cần chứng minh \(\dfrac{\left(a+b\right)^2}{a+b-2}\ge8\)
\(\Leftrightarrow\left(a+b\right)^2\ge8\left(a+b-2\right)\)
\(\Leftrightarrow a^2+2ab+b^2\ge8a+8b-16\)
\(\Leftrightarrow\left(a+b-4\right)^2\ge0\) luôn đúng
Thay A,B,C vào vế trái , ta có :
\(VT=\left(1+\frac{x-y}{x+y}\right)\left(1+\frac{y-z}{y+z}\right)\left(1+\frac{z-x}{z+x}\right)\)
\(=\left(\frac{x+y+x-y}{x+y}\right)\left(\frac{y+z+y-z}{y+z}\right)\left(\frac{z+x+z-x}{z+x}\right)\)
\(=\frac{2x}{x+y}.\frac{2y}{y+z}.\frac{2z}{z+X}\) \(=\frac{8xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)} \) (1)
Thay A,B,C vào vế phải , ta có
\(VP=\left(1-\frac{x-y}{x+y}\right)\left(1-\frac{y-z}{y+z}\right)\left(1-\frac{z-x}{z+x}\right)\)
\(=\left(\frac{x+y-x+y}{x+y}\right)\left(\frac{y+z-y+z}{y+z}\right)\left(\frac{z+x-z+x}{z+x}\right)\)
\(=\frac{2y}{x+y}.\frac{2z}{y+z}.\frac{2x}{z+x}=\frac{8xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\) (2)
Từ (1),(2) => đpcm
a,\(\dfrac{x-y}{xy}+\dfrac{y-z}{yz}+\dfrac{z-x}{zx}\)
=\(\dfrac{\left(x-y\right).z}{xyz}+\dfrac{\left(y-z\right).x}{xyz}+\dfrac{\left(z-x\right).y}{xyz}\)
=\(\dfrac{xz-yz}{xyz}+\dfrac{xy-xz}{xyz}+\dfrac{yz-xy}{xyz}\)
=\(\dfrac{xz-yz+xy-xz+yz-xy}{xyz}\)
=\(\dfrac{0}{xyz}\)=0
Vậy biểu thức trên ko phụ thuộc vào x,y,z
b,\(\dfrac{1}{\left(x-y\right).\left(y-z\right)}-\dfrac{1}{\left(x-z\right).\left(y-z\right)}-\dfrac{1}{\left(x-y\right).\left(x-z\right)}\)
=\(\dfrac{1.\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}-\dfrac{\left(x-y\right).1}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}-\dfrac{1\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
=\(\dfrac{x-z-x+y-y+z}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=\(\dfrac{0}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)=0
Vậy biểu thức trên ko phụ thuộc vào x,y,z
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)