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TỪ GT => \(3\le xy+yz+zx\)
=> \(P\ge\frac{x^3}{\sqrt{y^2+xy+yz+zx}}+\frac{y^3}{\sqrt{z^2+xy+yz+zx}}+\frac{z^3}{\sqrt{x^2+xy+yz+zx}}\)
=> \(P\ge\frac{x^3}{\sqrt{\left(x+y\right)\left(y+z\right)}}+\frac{y^3}{\sqrt{\left(z+x\right)\left(z+y\right)}}+\frac{z^3}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
TA ÁP DỤNG BĐT CAUCHY 2 SỐ SẼ ĐƯỢC:
=> \(\hept{\begin{cases}\sqrt{x+y}.\sqrt{y+z}\le\frac{x+2y+z}{2}\\\sqrt{z+x}.\sqrt{z+y}\le\frac{x+y+2z}{2}\\\sqrt{x+y}.\sqrt{x+z}\le\frac{2x+y+z}{2}\end{cases}}\)
=> \(P\ge\frac{2x^3}{x+2y+z}+\frac{2y^3}{x+y+2z}+\frac{2z^3}{2x+y+z}\)
=> \(P\ge\frac{2x^4}{x^2+2xy+2xz}+\frac{2y^4}{xy+y^2+2yz}+\frac{2z^4}{2xz+yz+z^2}\)
TA TIẾP TỤC ÁP DỤNG BĐT CAUCHY - SCHWARZ SẼ ĐƯỢC:
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)
TA CÓ 1 BĐT SAU: \(xy+yz+zx\le x^2+y^2+z^2\) (*)
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}\)
=> \(P\ge\frac{2\left(x^2+y^2+z^2\right)^2}{4\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{2}\)
TA LẠI 1 LẦN NỮA SỬ DỤNG BĐT (*) SẼ ĐƯỢC:
=> \(P\ge\frac{xy+yz+zx}{2}\ge\frac{3}{2}\left(gt\right)\)
DẤU "=" XẢY RA <=> \(x=y=z\)
VẬY P MIN \(=\frac{3}{2}\Leftrightarrow x=y=z=1\)
Ta có :
\(P\ge\frac{x^3}{\sqrt{y^2+xy+yz+zx}}+\frac{y^3}{\sqrt{z^2+xy+yz+zx}}+\frac{z^3}{\sqrt{z^2+xy+yz+zx}}\)
\(=\frac{x^3}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{y^3}{\sqrt{\left(z+x\right)\left(z+y\right)}}+\frac{z^3}{\sqrt{\left(x+y\right)\left(x+z\right)}}\)
\(\ge\frac{2x^3}{x+2y+z}+\frac{2y^3}{x+y+2z}+\frac{2z^3}{2x+y+z}\)\(\ge2.\frac{\left(x^2+y^2+z^2\right)^2}{\left(x^2+y^2+z^2\right)+3.\left(xy+yz+zx\right)}\ge2.\frac{\left(xy+yz+zx\right)^2}{4.\left(xy+yz+zx\right)}\ge2.\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
\(\frac{\left(x+y+z\right)^2}{3}\ge xy+yz+zx\Rightarrow x+y+z\ge3\)
\(P=\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}+\frac{y^2}{\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}}+\frac{z^2}{\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}+\sqrt{\left(y+2\right)\left(y^2-2y+4\right)}+\sqrt{\left(z+2\right)\left(z^2-2z+4\right)}}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x+2+x^2-2x+4\right)+\left(y+2+y^2-2y+4\right)+\left(z+2+z^2-2z+4\right)}\)
\(\Rightarrow P\ge\frac{2\left(x+y+z\right)^2}{\left(x^2+y^2+z^2\right)-\left(x+y+z\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)-2\left(xy+yz+zx\right)+18}=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-\left(x+y+z\right)+12}\)
Dự đoán Min P=1 khi x+y+z=3
Đặt \(t=x+y+z\ge3\)
\(\Rightarrow P\ge\frac{2t^2}{t^2-t+12}\Rightarrow P-1\ge\frac{t^2+t-12}{t^2-t+12}=\frac{\left(t-3\right)\left(t+4\right)}{t^2-t+12}\ge0\)
\(\Rightarrow P\ge1\)
Áp dụng bđt bunhiacopxki, ta có:
\(\left(x^2+\frac{1}{x^2}\right)\left(1+16\right)\ge\left(x+\frac{4}{x}\right)^2\) => \(x^2+\frac{1}{x^2}\ge\frac{\left(x+\frac{4}{x}\right)^2}{17}\)
=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{x+\frac{4}{x}}{\sqrt{17}}=\frac{x}{\sqrt{17}}+\frac{4}{x\sqrt{17}}\)
CMTT: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{y}{\sqrt{17}}+\frac{4}{\sqrt{17}y}\)
\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{z}{\sqrt{17}}+\frac{4}{\sqrt{17}z}\)
=> A \(\ge\frac{x+y+z}{\sqrt{17}}+\frac{4}{\sqrt{17}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{x+y+z}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}\)(bđt: 1/a + 1/b + 1/c > = 9/(a+b+c)
=> A \(\ge\frac{16\left(x+y+z\right)}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}-\frac{15\left(x+y+z\right)}{\sqrt{17}}\)
A \(\ge2\sqrt{\frac{16\left(x+y+z\right)}{\sqrt{17}}\cdot\frac{36}{\sqrt{17}\left(x+y+z\right)}}-\frac{15\cdot\frac{3}{2}}{\sqrt{17}}\)(Bđt cosi + bđt: x + y + z < = 3/2)
A \(\ge\frac{48}{\sqrt{17}}-\frac{45}{2\sqrt{17}}=\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra <=> x = y= z = 1/2
Vậy MinA = \(\frac{3\sqrt{17}}{2}\) <=> x = y = z = 1/2
tth, Phạm Minh Quang, Lê Thị Hồng Vân, Thiên Thảo, Sky SơnTùng, @Trần Thanh Phương, @nguyễn thị ngọc thơ, @Nguyễn Việt Lâm, @Akai Haruma
Giúp e vs ạ!