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\(\dfrac{1}{\left(a+b+a+c\right)^2}\le\dfrac{1}{4\left(a+b\right)\left(a+c\right)}=\dfrac{1}{4\left(a^2+ab+bc+ca\right)}\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\)
\(\le\dfrac{1}{64}\left(\dfrac{1}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=\dfrac{1}{64}\left(\dfrac{2}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Tương tự và cộng lại:
\(P\le\dfrac{1}{64}\left(\dfrac{4}{a^2}+\dfrac{4}{b^2}+\dfrac{4}{c^2}\right)=\dfrac{1}{16}.3=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bđt: \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(1\right)\)
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
\(\Rightarrow P\le\dfrac{1}{16}\left[\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)^2+\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)^2+\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)^2\right]\)\(\Rightarrow16P\le\dfrac{2}{\left(a+b\right)^2}+\dfrac{2}{\left(b+c\right)^2}+\dfrac{2}{\left(a+c\right)^2}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(a+b\right)\left(b+c\right)}+\dfrac{2}{\left(b+c\right)\left(c+a\right)}\)
Áp dụng: \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\) với a+b=x,b+c=y,c+a=z
\(\Rightarrow16P\le\dfrac{4}{\left(a+b\right)^2}+\dfrac{4}{\left(b+c\right)^2}+\dfrac{4}{\left(c+a\right)^2}\)
Ta có: \(\dfrac{1}{\left(a+b\right)^2}\le4.16.\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\)(do (1))
\(\Rightarrow16P\le\dfrac{1}{4}.16\left[\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)^2+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)^2\right]=\dfrac{1}{4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\right)\le\dfrac{1}{4}.4.\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=3\)(do(2) và \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=3\))
\(\Rightarrow P\le\dfrac{3}{16}\)
\(ĐTXR\Leftrightarrow a=b=c=1\)
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Từ \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)+2017\)
\(\Leftrightarrow7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2017\)\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\le2017\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(T=\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
\(=\dfrac{1}{\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{\left(2+1\right)\left(2c^2+a^2\right)}}\)
\(\le\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}\le\dfrac{1}{9}\left(\dfrac{2^2}{2a}+\dfrac{1^2}{b}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2b}+\dfrac{1^2}{c}\right)+\dfrac{1}{9}\left(\dfrac{2^2}{2c}+\dfrac{1^2}{a}\right)\)
\(\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)\)\(=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\le\sqrt{\left(\dfrac{1}{81}+\dfrac{1}{81}+\dfrac{1}{81}\right)\left(\dfrac{9}{a^2}+\dfrac{9}{b^2}+\dfrac{9}{c^2}\right)}\)
\(\le\sqrt{\dfrac{1}{81}\cdot3\cdot9\cdot2017}=\sqrt{\dfrac{2017}{3}}\)
Vậy \(T_{Max}=\sqrt{\dfrac{2017}{3}}\) khi \(a=b=c=\sqrt{\dfrac{3}{2017}}\)
So kimochiii~
Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)
\(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)
\(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)
Cộng 3 cái vào, ta có
A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)
Vậy A min = 24
Neetkun ^^
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^3(b+c)}+\frac{a(b+c)}{4}\geq 2\sqrt{\frac{1}{a^3(b+c)}.\frac{a(b+c)}{4}}=2\sqrt{\frac{1}{4a^2}}=\frac{1}{a}=\frac{abc}{a}=bc\)
Tương tự:
\(\frac{1}{b^3(c+a)}+\frac{b(c+a)}{4}\geq \frac{1}{b}=ac\)
\(\frac{1}{c^3(a+b)}+\frac{c(a+b)}{4}\geq \frac{1}{c}=ab\)
Cộng theo vế:
\(\Rightarrow \text{VT}+\frac{ab+bc+ac}{2}\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{ab+bc+ac}{2}\)
Tiếp tục áp dụng AM-GM: \(ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}=3\)
\(\Rightarrow \text{VT}\ge \frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi $a=b=c=1$
Lời giải:
Đặt vế trái là $A$
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)(a+b+b+c+c+c)\geq (1+1+1+1+1+1)^2\)
\(\Leftrightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{36}{a+2b+3c}\)
Hoàn toàn TT:
\(\frac{1}{b}+\frac{2}{c}+\frac{3}{a}\geq \frac{36}{b+2c+3a}\)
\(\frac{1}{c}+\frac{2}{a}+\frac{3}{b}\geq \frac{36}{c+2a+3b}\)
Cộng theo vế:
\(\Rightarrow 6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 36A\)
\(\Rightarrow A\leq \frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo đkđb: \(ab+bc+ac=abc\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Do đó: \(A\leq \frac{1}{6}< \frac{3}{16}\) (đpcm)
Áp dụng BĐT Cô-si cho các số dương ta có:
(2a+b+c)2 = \(\left[\left(a+b\right)+\left(a+c\right)\right]^2\) \(\ge\) 4(a+b)(a+c)
\(\Rightarrow\) \(\dfrac{1}{\left(2a+b+c\right)^2}\) \(\le\) \(\dfrac{1}{4\left(a+b\right)\left(a+c\right)}\)
Tương tự : \(\dfrac{1}{\left(2b+c+a\right)^2}\) \(\le\) \(\dfrac{1}{4\left(b+c\right)\left(b+a\right)}\)
\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\) \(\dfrac{1}{4\left(c+b\right)\left(c+a\right)}\)
Cộng theo vế 3 đẳng thức trên
\(\dfrac{1}{\left(2a+b+c\right)^2}\)+\(\dfrac{1}{\left(2b+c+a\right)^2}\)+\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\)\(\dfrac{1}{4}\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(b+c\right)\left(b+a\right)}+\dfrac{1}{\left(c+b\right)\left(c+a\right)}\right)\)
=\(\dfrac{1}{4}\left(\dfrac{b+c+a+b+c+a}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right)\)
Áp dụng BĐT Cô-si ta có:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
\(\Rightarrow\) P \(\le\) \(\dfrac{a+b+c}{16abc}\) = \(\dfrac{1}{16}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\) \(\le16\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\) = \(\dfrac{3}{16}\)
\(\Rightarrow\) Pmax = \(\dfrac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow\) a = b = c = 1
Vậy Pmax = \(\dfrac{3}{16}\) \(\Leftrightarrow\) a = b = c = 1
Xí câu BĐT:
ta cần chứng minh \(\dfrac{a^2}{b^2c}+\dfrac{b^2}{c^2a}+\dfrac{c^2}{a^2b}\ge\dfrac{ab+bc+ca}{abc}\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
Áp dụng BĐT cauchy:
\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}=2a^2\)
tương tự ta có:\(\dfrac{b^3}{c}+bc\ge2b^2;\dfrac{c^3}{a}+ac\ge2c^2\)
cả 2 vế các BĐT đều dương,cộng vế với vế ta có:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)\)
mà a2+b2+c2\(\ge ab+bc+ca\) ( chứng minh đầy đủ nhá)
do đó \(S=\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(ab+bc+ca\right)-ab+bc+ca=ab+bc+ca\)
suy ra BĐT ban đầu đúng
dấu = xảy ra khi và chỉ khi a=b=c.
P/s: cách khác :Áp dụng BĐT cauchy-schwarz:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\)
\(S\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Câu hệ này =))
b, Từ hệ đã cho ta thấy x,y > 0
Trừ vế cho vế pt (1) và (2) của hệ ta được:
\(x^4-y^4=4y-4x\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)=4\left(y-x\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)+4\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)\left(x^2+y^2\right)+4\right]=0\)
\(\Leftrightarrow x-y=0\) ( Vì \(\left(x+y\right)\left(x^2+y^2\right)+4>0\) với x,y > 0)
\(\Leftrightarrow x=y\)
Với x = y thay vào pt đầu của hệ ta được:
\(x^4-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x-1=0\) ( Vì \(x^2+2x+3>0\) )
\(\Leftrightarrow x=1\)
Với x=1 suy ra y=1
Vậy hệ đã cho có nghiệm duy nhất (x;y) = (1;1)
ko biết mk làm có đúng ko nhma có gì sai thì đừng trách mk nhé
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge\dfrac{63}{a^2+b^2+c^2}\)
\(6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{a}{ac}\right)+2021\ge\dfrac{54}{ab+bc+ac}+2021\ge\dfrac{54}{a^2+b^2+c^2}+2021\)
<=>\(\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{2021}{9}\)
\(p^2=\left(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\right)^2\)
áp dụng bđt \(a^2+b^2+c^2\ge\dfrac{1}{3}\left(a+b+c\right)^2\)
\(p^2\le3.\left(\dfrac{1}{3\left(2a^2+b^2\right)}+\dfrac{1}{3\left(2b^2+c^2\right)}+\dfrac{1}{3\left(2c^2+a^2\right)}\right)=\dfrac{1}{2a^2+b^2}+\dfrac{1}{2b^2+c^2}+\dfrac{1}{2c^2+a^2}\)
\(< =>p^2\le\dfrac{9}{2a^2+b^2+2b^2+c^2+2c^2+a^2}\)
<=> \(p^2\le3.\dfrac{1}{a^2+b^2+c^2}=\dfrac{2021}{3}< =>p\le\sqrt{\dfrac{2021}{3}}\)
dấu bằng xảy ra khi \(a=b=c=\sqrt{\dfrac{3}{2021}}\)
\(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2021\le6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2021\)
\(\Rightarrow2021\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\sqrt{2021.3}=\sqrt{6063}\)
Từ đó:
\(\sqrt{3\left(2a^2+b\right)}=\sqrt{\left(2+1\right)\left(2a^2+b^2\right)}\ge\sqrt{\left(2a+b\right)^2}=2a+b\)
\(\Rightarrow\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}\le\dfrac{1}{2a+b}=\dfrac{1}{a+a+b}\le\dfrac{1}{9}\left(\dfrac{2}{a}+\dfrac{1}{b}\right)\)
Tương tự: \(\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{b}+\dfrac{1}{c}\right)\) ; \(\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\le\dfrac{1}{9}\left(\dfrac{2}{c}+\dfrac{1}{a}\right)\)
Cộng vế:
\(\Rightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\dfrac{1}{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{\sqrt{6063}}{3}\)
\(P_{max}=\dfrac{\sqrt{6063}}{3}\) khi \(a=b=c=\dfrac{3}{\sqrt{6063}}\)
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(2a+b+c=(a+b)+(a+c)\geq 2\sqrt{(a+b)(a+c)}\)
\(\Rightarrow (2a+b+c)^2\geq 4(a+b)(a+c)\)
\(\Rightarrow \frac{1}{(2a+b+c)^2}\leq \frac{1}{4(a+b)(a+c)}\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(P\leq \frac{1}{4}\left(\frac{1}{(a+b)(a+c)}+\frac{1}{(b+c)(b+a)}+\frac{1}{(c+a)(c+b)}\right)\)
\(\Leftrightarrow P\leq \frac{1}{4}.\frac{(b+c)+(c+a)+(a+b)}{(a+b)(b+c)(c+a)}\)
\(\Leftrightarrow P\leq \frac{a+b+c}{2(a+b)(b+c)(c+a)}\)
Lại có: \((a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ac}=8abc\) (theo AM-GM)
\(\Rightarrow P\leq \frac{a+b+c}{2.8abc}=\frac{a+b+c}{16abc}(1)\)
Tiếp tục áp dụng BĐT AM-GM:
\(\frac{1}{a^2}+\frac{1}{b^2}\geq \frac{2}{ab}; \frac{1}{b^2}+\frac{1}{c^2}\geq \frac{2}{bc}; \frac{1}{c^2}+\frac{1}{a^2}\geq \frac{2}{ac}\)
\(\Rightarrow 2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\geq 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Leftrightarrow 3\geq \frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a+b+c}{abc}\)
\(\Rightarrow a+b+c\leq 3abc(2)\)
Từ \((1); (2)\Rightarrow P\leq \frac{3abc}{16abc}=\frac{3}{16}\)
Vậy \(P_{\max}=\frac{3}{16}\). Dấu bằng xảy ra khi \(a=b=c=1\)