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1.Ta có :\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^2-xy+y^2\) (do x+y=1)
\(=\dfrac{3}{4}\left(x-y\right)^2+\dfrac{1}{4}\left(x+y\right)^2\ge\dfrac{1}{4}\left(x+y\right)^2\)\(=\dfrac{1}{4}.1=\dfrac{1}{4}\)
Dấu "=" xảy ra khi :\(x=y=\dfrac{1}{2}\)
Vậy \(x^3+y^3\ge\dfrac{1}{4}\)
2.
a) Sửa đề: \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\ge0\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng vì \(a,b\ge0\))
Đẳng thức xảy ra \(\Leftrightarrow a=b\)
b) Lần trước mk giải rồi nhá
3.
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel\(P=\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}\ge\dfrac{\left(1+1+1\right)^2}{\left(x+y+z\right)+3}=\dfrac{9}{3+3}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}=\dfrac{1}{y+1}=\dfrac{1}{z+1}\\x+y+z=3\end{matrix}\right.\Leftrightarrow x=y=z=1\)
b) \(Q=\dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1}\le\dfrac{x}{2\sqrt{x^2.1}}+\dfrac{y}{2\sqrt{y^2.1}}+\dfrac{z}{2\sqrt{z^2.1}}\)
\(=\dfrac{x}{2x}+\dfrac{y}{2y}+\dfrac{z}{2z}=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\)
Đẳng thức xảy ra \(\Leftrightarrow x^2=y^2=z^2=1\Leftrightarrow x=y=z=1\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\) (1)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+ayz+bxz}{abc}\right)=1\)
Kết hợp với (1) ta có đpcm
Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow ayz+bxz+cxy=0\left(1\right)\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xyc+ayz+xbz}{abc}\right)=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)(đpcm)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\Leftrightarrow ayz+bxz+cxy=0\)
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ac}\right)\)
\(=\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2-2\left(\dfrac{cxy+ayz+bzx}{abc}\right)\)\(=1-0=1\left(dpcm\right)\)
bài 3:
a, đặt x12=y9=z5=kx12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
Bài 1:
Áp dụng BĐT AM-GM cho các số thực dương ta có:
\(\frac{x^2}{y+z}+\frac{y+z}{4}\geq 2\sqrt{\frac{x^2}{4}}=x\)
\(\frac{y^2}{x+z}+\frac{x+z}{4}\geq 2\sqrt{\frac{y^2}{4}}=y\)
\(\frac{z^2}{x+y}+\frac{x+y}{4}\geq 2\sqrt{\frac{z^2}{4}}=z\)
Cộng theo vế:
\(\Rightarrow M+\frac{y+z}{4}+\frac{x+z}{4}+\frac{x+y}{4}\geq x+y+z\)
\(\Leftrightarrow M\geq \frac{x+y+z}{2}=\frac{2}{2}=1\)
Vậy GTNN của $M$ là $1$. Đẳng thức xảy ra tại $x=y=z=\frac{2}{3}$
Bài 2:
\(\text{VT}=(a+1)-\frac{b^2(a+1)}{b^2+1}+(b+1)-\frac{c^2(b+1)}{c^2+1}+(c+1)-\frac{a^2(c+1)}{a^2+1}\)
\(=(a+b+c+3)-\left(\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\right)\)
\(=6-M(*)\)
Xét \(M=\frac{b^2(a+1)}{b^2+1}+\frac{c^2(b+1)}{c^2+1}+\frac{a^2(c+1)}{a^2+1}\). Áp dụng BĐT AM-GM:
\(M\leq \frac{b^2(a+1)}{2b}+\frac{c^2(b+1)}{2c}+\frac{a^2(c+1)}{2a}=\frac{ab+bc+ac+a+b+c}{2}=\frac{ab+bc+ac+3}{2}\)
\(\leq \frac{\frac{(a+b+c)^2}{3}+3}{2}=3(**)\)
Từ \((*); (**)\Rightarrow \text{VT}=6-M\geq 6-3=3\) (đpcm)
Dấu "=" xảy ra khi $a=b=c=1$
bài 2
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{a+c}{b}=>a=b=c\)
=>x=2 => A=(4-2+1)^2016 =3^2016