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\(\frac{1}{\sqrt[3]{x+3y}}\ge\frac{1}{\frac{x+3y+1+1}{3}}=\frac{3}{x+3y+2}\\ \text{Tương tự }\Rightarrow P\ge\frac{3}{x+3y+2}+\frac{3}{y+3z+2}+\frac{3}{z+3x+2}\\ \ge3\cdot\frac{9}{x+3y+2+y+3z+2+z+3x+2}\\ =3\)
Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)(với a,b,c > 0 )
\(\Leftrightarrow a^3+b^3+c^3\ge3abc\Leftrightarrow abc\le\frac{a^3+b^3+c^3}{3}\).
AD CT trên ta có :
\(1.1.\sqrt[3]{x+3y}\le\frac{1+1+x+3y}{3}\Leftrightarrow\sqrt[3]{x+3y}\le\frac{x+3y+2}{3}\).
Cmtt có : \(\sqrt[3]{y+3z}\le\frac{y+3z+2}{3};\sqrt[3]{z+3x}\le\frac{z+3x+2}{3}\)
\(\Rightarrow\sqrt[3]{x+3y}+\sqrt[3]{y+3z}+\sqrt[3]{z+3x}\le\frac{4\left(x+y+z\right)+6}{3}=3\)
AD BĐT Cộng mẫu số ta có:
\(\frac{1}{\sqrt[3]{x+3y}}+\frac{1}{\sqrt[3]{y+3z}}+\frac{1}{\sqrt[3]{z+3x}}\ge\frac{\left(1+1+1\right)^2}{\sqrt[3]{x+3y}+\sqrt[3]{y+3z}+\sqrt[3]{z+3x}}\ge\frac{9}{3}=3\)Dấu ''='' xảy ra \(\Leftrightarrow a=b=c=\frac{1}{4}\)
Vậy GTNN của b.thức là P = 3 khi a = b = c =\(\frac{1}{4}\)
Đặt vế trái là P, ta có:
\(P\le\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\)
Nên ta chỉ cần chứng mình: \(\sqrt{3\left(\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\right)}\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x}{z+3x}+\dfrac{y}{x+3y}+\dfrac{z}{y+3z}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{3x}{z+3x}-1+\dfrac{3y}{x+3y}-\dfrac{3z}{y+3z}-1\le\dfrac{9}{4}-3\)
\(\Leftrightarrow\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}\ge\dfrac{3}{4}\)
BĐT trên đúng do:
\(\dfrac{z}{z+3x}+\dfrac{x}{x+3y}+\dfrac{y}{y+3z}=\dfrac{z^2}{z^2+3zx}+\dfrac{x^2}{x^2+3xy}+\dfrac{y^2}{y^2+3yz}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+xy+yz+zx}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\dfrac{1}{3}\left(x+y+z\right)^2}=\dfrac{3}{4}\)
1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Chắc bạn ghi nhầm căn thức thứ 2
\(A2\sqrt{2}=2\sqrt{\left(2x+4\right)\left(x^2-2x+4\right)}+2\sqrt{\left(2y+4\right)\left(y^2-2y+4\right)}+2\sqrt{\left(2z+4\right)\left(z^2-2z+4\right)}\)
\(A2\sqrt{2}\le2x+4+x^2-2x+4+2y+4+y^2-2y+4+2z+4+z^2-2z+4\)
\(A2\sqrt{2}\le x^2+y^2+z^2+24=72\)
\(A\le18\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=z=4\)
\(VT=\sqrt[3]{1.1.\left(x+3y\right)}+\sqrt[3]{1.1.\left(y+3z\right)}+\sqrt[3]{1.1.\left(z+3x\right)}\)
\(VT\le\frac{1}{3}\left(1+1+x+3y\right)+\frac{1}{3}\left(1+1+y+3z\right)+\frac{1}{3}\left(1+1+z+3x\right)\)
\(VT\le\frac{1}{3}\left(6+4\left(x+y+z\right)\right)=3\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
Dấu = k xảy ra vì nếu x=y=z=\(\frac{1}{3}\) thì k thỏa mãn đk đề bài.