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Áp dụng BDT cauchy-schwarz:
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}\ge\dfrac{\left(a+b\right)^2}{2ab+c\left(a+b\right)}\ge\dfrac{\left(a+b\right)^2}{\dfrac{\left(a+b\right)^2}{2}+c\left(a+b\right)}=\dfrac{2}{1+\dfrac{2c}{a+b}}\)
Đặt \(\sqrt{\dfrac{2c}{a+b}}=x\) thì \(VT\ge\dfrac{2}{1+x^2}+x\)
Tiếp tục áp dụng BĐT AM-GM:
\(\dfrac{2}{1+x^2}+x=2-\dfrac{2x^2}{1+x^2}+x\ge2-\dfrac{2x^2}{2x}+x=2\)
Dấu = xảy ra khi x=1 và a=b hay a=b=c
x=ay=bz
\(\sqrt{xy}=\dfrac{1}{\sqrt{a}}\sqrt{x.ay}\le\dfrac{1}{2\sqrt{a}}\left(x+ay\right)\)
\(3\sqrt{yz}=\dfrac{3}{\sqrt{ab}}\sqrt{ay.bz}\le\dfrac{3}{2\sqrt{ab}}\left(ay+bz\right)\)
\(5\sqrt{xz}=\dfrac{5}{\sqrt{b}}.\sqrt{x.bz}\le\dfrac{5}{2\sqrt{b}}\left(x+bz\right)\)
\(\Rightarrow VF\le x\left(\dfrac{1}{2\sqrt{a}}+\dfrac{5}{2\sqrt{b}}\right)+y\left(\dfrac{\sqrt{a}+3\sqrt{a}}{2}\right)+z\left(\dfrac{3\sqrt{b}+5\sqrt{b}}{2}\right)\)
\(=x\left(\dfrac{1}{2\sqrt{a}}+\dfrac{5}{2\sqrt{b}}\right)+y.2\sqrt{a}+z.4\sqrt{b}\)
\(\Rightarrow\dfrac{4\sqrt{b}}{4}=\dfrac{2\sqrt{a}}{2}=\dfrac{\sqrt{b}+5\sqrt{a}}{6\sqrt{ab}}\Rightarrow a=b=1\)
troll dữ vậy .-.
Ta có: \(Q=2\left(1-y-z\right)+\left(y-z\right)^2+4\sqrt{yz}\)
\(=2+\left(\sqrt{y}-\sqrt{z}\right)^2\left(y+z+2\sqrt{yz}-2\right)\)
\(=2+\left(\sqrt{y}-\sqrt{z}\right)^2\left[-\left(\sqrt{y}-\sqrt{z}\right)^2-2x\right]\le2\)
Dấu = xảy ra khi y=z
Viết lại bài toán: Cho \(a^2+b^2+c^2=1\). Tìm max \(\sum\dfrac{a}{b^2+c^2}\)
với a=2x, b=3y, c=4z.
Áp dụng BĐT AM-GM:
\(a\left(b^2+c^2\right)=\dfrac{1}{\sqrt{2}}\sqrt{2a^2\left(1-a^2\right)\left(1-a^2\right)}\le\dfrac{1}{\sqrt{2}}\sqrt{\dfrac{8}{27}}=\dfrac{2}{3\sqrt{3}}\)
Do đó \(VT\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Vậy \(A_{Min}=\dfrac{3\sqrt{3}}{2}\)
Từ GT ta có: \(3=\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
Suy ra \(3\le x+y+z\)
Áp dụng AM-GM:
\(VT\le\dfrac{x^2}{2x^2\sqrt{yz}}+\dfrac{y^2}{2y^2\sqrt{xz}}+\dfrac{z^2}{2z^2\sqrt{xy}}=\dfrac{1}{2}\sum\dfrac{1}{\sqrt{xy}}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2\sqrt{xyz}}\le\dfrac{\sqrt{3\left(x+y+z\right)}}{2\sqrt{xyz}}\le\dfrac{1}{2}\sqrt{\dfrac{\left(x+y+z\right)^2}{xyz}}\)
\(\le\dfrac{1}{2}\sqrt{\dfrac{3\left(x^2+y^2+z^2\right)}{xyz}}=\dfrac{3}{2}\)
Vậy \(P_{Max}=\dfrac{3}{2}\)
Viết lại BĐT:\(\dfrac{a^2b}{a^2b+2}+\dfrac{b^2c}{b^2c+2}+\dfrac{c^2a}{c^2a+2}\le1\)
\(VT\le\sum\dfrac{a^2b}{3\sqrt[3]{a^4b^2}}=\dfrac{1}{3}\left(\sqrt[3]{a^2b}+\sqrt[3]{b^2c}+\sqrt[3]{c^2a}\right)\)
\(\le\dfrac{1}{9}\left(3a+3b+3c\right)=1\)
Suy ra đpcm
\(P\ge12\sqrt[12]{\dfrac{a^3.b^4.c^5}{a^9.b^8.c^7}}=12\sqrt[12]{\dfrac{1}{a^6.b^4.c^2}}=12\sqrt[12]{\dfrac{1}{a^2.a^2.a^2.b^2.b^2.c^2}}\)
\(\ge12\sqrt[12]{\dfrac{1}{\dfrac{\left(3a^2+2b^2+c^2\right)^6}{6^6}}}\ge12\sqrt{6}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{6}}\)
Khai triển Abel ta có:
\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)
\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)
\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)
\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)
Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)
\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)
\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).
\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)
Dấu = xảy ra khi x=y=z=1 hay a=b=c
Áp dụng BĐT Bunyakovsky:
\(\left(1+xy\right)\left(1+\dfrac{y}{x}\right)\ge\left(1+y\right)^2\)\(\Rightarrow\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{\dfrac{x}{x+y}}{1+xy}\)
\(\left(1+xy\right)\left(1+\dfrac{x}{y}\right)\ge\left(1+x\right)^2\Rightarrow\dfrac{1}{\left(1+x\right)^2}\ge\dfrac{\dfrac{y}{x+y}}{1+xy}\)
Cộng theo vế ta thu được đpcm.Dấu = chỉ xảy ra khi x=y=1