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\(\left\{{}\begin{matrix}x;y>0\\x+y=1\end{matrix}\right.\)\(\Rightarrow0< xy=t\le\dfrac{1}{4}\)
\(x^4+y^4=\left(1-2t\right)^2-2t\)
\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge5\Leftrightarrow A=8\left[\left(1-2t\right)^2-2t\right]+\dfrac{1}{t}-5\ge0\)
\(\Leftrightarrow16t^2-32t+\dfrac{1}{t}+3\ge0\)\(\Leftrightarrow16t^3-32t^2+3t+1\ge0\)
<=>\(16t^3-4t^2-28t^2+7t-4t+1\ge0\)
\(\Leftrightarrow4t^2\left(4t-1\right)-7t\left(4t-1\right)-\left(4t-1\right)\ge0\)
\(\Leftrightarrow\left(4t-1\right)\left(4t^2-7t-1\right)\ge0\)
\(\Leftrightarrow B=\left(4t-1\right)\left(8t-7-\sqrt{65}\right)\left(8t-7+\sqrt{65}\right)\ge0\)
\(0< t\le\dfrac{1}{4}\Rightarrow\)\(\left\{{}\begin{matrix}4t-1\le0\\8t-7+\sqrt{65}>0\\8t-7-\sqrt{5}< 0\end{matrix}\right.\) \(\Rightarrow B\ge0\)
mọi phép biến đổi <=> => dpcm
Sử dụng BĐT Cauchy-Schwarz nhiều lần, cộng với BĐT phụ \(\dfrac{1}{xy}\ge\dfrac{4}{\left(x+y\right)^2}\), ta có:
\(8\left(x^4+y^4\right)+\dfrac{1}{xy}\ge\dfrac{8\left(x^2+y^2\right)^2}{2}+\dfrac{4}{\left(x+y\right)^2}=4\left(x^2+y^2\right)^2+4\ge4\left[\dfrac{\left(x+y\right)^2}{2}\right]^2+4=5\)
Đẳng thức xảy ra khi \(x=y=\dfrac{1}{2}\)
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)
\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)
\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)
\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)
\(A=8\left(x^4+y^4\right)+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\ge8\left(x^4+y^4\right)+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2\left(x^2+y^2\right)}+\frac{1}{2xy}\)
\(\Rightarrow A\ge8\left(x^4+y^4\right)+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}+\frac{1}{2\left(\frac{x+y}{2}\right)^2}\)
\(\Rightarrow A\ge3\sqrt[3]{8\left(x^4+y^4\right)\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}\cdot\frac{1}{2\sqrt{2\left(x^4+y^4\right)}}}+\frac{1}{2\cdot\frac{1}{4}}=3+2=5\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
\(xy\ne0,x,y\ne1\)
\(A=\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x+y\right)}{x^2y^2+3}\)
\(xét:\dfrac{2\left(x+y\right)}{x^2y^2+3}=\dfrac{2}{x^2y^2+3}\left(1\right)\)
\(\dfrac{x^{ }}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{x^4-x-y^4+y}{\left(x^3-1\right)\left(y^3-1\right)}\left(2\right)\)
\(xét:\) \(x^4-x-y^4+y=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3-1\right)\)
\(=\left(x-y\right)\left[\left(x+y\right)^3-3xy\left(x+y\right)+xy\left(x+y\right)-1\right]\)
\(=\left(x-y\right)\left(1-3xy+xy-1\right)\)
\(=\left(x-y\right)\left(-2xy\right)=-2xy\left(x-y\right)=2xy\)
\(xét\) \(\left(y^3-1\right)\left(x^3-1\right)=x^3y^3-\left[\left(x+y\right)^3-3xy\left(x+y\right)\right]+1\)
\(=x^3y^3-\left(1-3xy\right)+1=x^3y^3+3xy=xy\left(x^2y^2+3\right)\)
\(\Rightarrow\left(2\right)\Leftrightarrow\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)
\(\left(1\right)\left(2\right)\Rightarrow A=\dfrac{2}{x^2y^2+3}-\dfrac{2\left(x-y\right)}{x^2y^2+3}=\dfrac{2-2x+2y}{x^2y^2+3}\ne0\left(đề-sai\right)\)
có bđt: a²+b² ≥ (a+b)²/2 (*)
(*) <=> 2a²+2b² ≥ a²+b²+2ab <=> a²+b²-2ab ≥ 0 <=> (a-b)² ≥ 0 bđt đúng, dấu "=" khi a = b
- - -
ad (*) 2 lần liên tiếp:
x^4 + y^4 ≥ (x²+y²)²/2 ≥ [(x+y)²/2]²/2 = (x+y)^4 /8 = 1/8
=> 8(x^4 + y^4) ≥ 1 (*)
mặt khác, có bđt: (x-y)² ≥ 0 <=> x²+y² ≥ 2xy <=> x²+y²+2xy ≥ 4xy <=> (x+y)² ≥ 4xy
=> 1/xy ≥ 4/(x+y)² = 4 (**)
(*) + (**): 8(x^4 + y^4) + 1/xy ≥ 1+4 = 5 (đpcm) dấu "=" khi x = y = 1/2
\(x+y=1\ge2\sqrt{xy}\Leftrightarrow xy\le\frac{1}{4}\)
\(A=8\left(x^4+y^4\right)+\frac{1}{xy}\ge16x^2y^2+\frac{1}{xy}=16x^2y^2+\frac{1}{4xy}+\frac{1}{4xy}+\frac{1}{2xy}\ge3\sqrt[3]{16x^2y^2.\frac{1}{4xy}.\frac{1}{4xy}}+\frac{1}{2.\frac{1}{4}}=5\)
Dâu ' = ' xảy ra khi x =y = 1/2
Áp dụng BĐT Cô-si :
\(\frac{1}{xy}\ge\frac{1}{\frac{\left(x+y\right)^2}{4}}\ge\frac{1}{\frac{1}{4}}=4\)
Do đó BĐT cần chứng minh \(\Leftrightarrow8\left(x^4+y^4\right)+4\ge5\)
Ta cần chứng minh BĐT sau là đủ : \(8\left(x^4+y^4\right)\ge1\)
Thật vậy: Áp dụng BĐT Cô-si :
\(x^4+\frac{1}{16}\ge\frac{x^2}{2};y^4+\frac{1}{16}\ge\frac{y^2}{2}\)
Cộng vế : \(x^4+y^4+\frac{1}{8}\ge\frac{x^2+y^2}{2}\ge\frac{\frac{\left(x+y\right)^2}{2}}{2}\ge\frac{\frac{1}{2}}{2}=\frac{1}{4}\)
\(\Leftrightarrow x^4+y^4\ge\frac{1}{4}-\frac{1}{8}=\frac{1}{8}\)
\(\Leftrightarrow8\left(x^4+y^4\right)\ge1\)
Ta có đpcm.
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)